Calculus

Integration

Integration is the inverse operation of differentiation. Given the derivative of a function, Integration recovers the original function (up to an additive constant). It is a central operation in calculus with wide-ranging applications in geometry, physics, and probability.

Two principal forms exist:

Indefinite integration (antidifferentiation): finds the general antiderivative of a function, producing a family of functions differing by a constant.
Definite integration: computes a numerical value representing the signed area under a curve between two limits.

These are linked by the Fundamental Theorem of Calculus, which unifies differentiation and Integration into a single coherent framework.

Notation

\int f(x)\,dx = F(x) + C

Where $F'(x) = f(x)$ and $C$ is the constant of integration. The symbol $\int$ is an elongated $S$ (for “sum”), $f(x)$ is the integrand, and $dx$ indicates the variable of integration.

1. Antiderivatives

Definition

A function $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x \in I$ . If $F$ is an antiderivative of $f$ Then so is $F + C$ for any constant $C \in \mathbb{R}$ .

Basic Rules

Power Rule

For $n \neq -1$ :

\int x^n\,dx = \frac\{x^\{n+1\}\}\{n+1\} + C

Constant Multiple Rule

For $k \in \mathbb{R}$ :

\int k f(x)\,dx = k \int f(x)\,dx

Sum and Difference Rule

\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx

Reciprocal Rule (Logarithmic)

\int \frac\{1\}\{x\}\,dx = \ln|x| + C

Common Antiderivatives

$f(x)$	$\int f(x)\,dx$
$k$	$kx + C$
$x^n$	$\dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
$\dfrac{1}{x}$	$\ln \\| x \\| + C$
$e^x$	$e^x + C$
$a^x$	$\dfrac{a^x}{\ln a} + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$
$\csc^2 x$	$-\cot x + C$
$\sec x \tan x$	$\sec x + C$
$\csc x \cot x$	$-\csc x + C$
$\tan x$	$\ln \\| \sec x \\| + C$
$\cot x$	$\ln \\| \sin x \\| + C$
$\frac{1}{\sqrt{1 - x^2}}$	$\arcsin x + C$
$\frac{1}{1 + x^2}$	$\arctan x + C$

Expand

Find $\int (3x^4 - 5x^2 + 7)\,dx$

\begin`\{aligned}` \int (3x^4 - 5x^2 + 7)\,dx &= 3 \cdot \frac\{x^5\}\{5\} - 5 \cdot \frac\{x^3\}\{3\} + 7x + C \\[6pt] &= \frac\{3x^5\}\{5\} - \frac\{5x^3\}\{3\} + 7x + C \end`\{aligned}`

Find $\int \left(\frac{2}{x} + 3e^x - \frac{1}{\cos^2 x}\right)\,dx$

\begin`\{aligned}` \int \left(\frac\{2\}\{x\} + 3e^x - \sec^2 x\right)\,dx &= 2\ln|x| + 3e^x - \tan x + C \end`\{aligned}`

2. Definite Integration

Definite Integral as a Limit of a Riemann Sum

For a function $f$ continuous on $[a, b]$ The definite integral is defined as:

\int_a^b f(x)\,dx = \lim_\{n \to \infty\} \sum_\{i=1\}^\{n\} f(x_i^*) \Delta x

Where $\Delta x = \dfrac{b - a}{n}$ and $x_i^*$ is a sample point in the $i$ -th subinterval.

Fundamental Theorem of Calculus — Part 1

If $f$ is continuous on $[a, b]$ and $F$ is defined by:

F(x) = \int_a^x f(t)\,dt

Then $F$ is differentiable on $(a, b)$ and:

F'(x) = f(x)

This establishes that differentiation and integration are inverse operations.

Fundamental Theorem of Calculus — Part 2 (Evaluation Theorem)

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ Then:

\int_a^b f(x)\,dx = F(b) - F(a)

This is written using bracket notation:

\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)

Area Under a Curve

If $f(x) \geq 0$ on $[a, b]$ Then $\int_a^b f(x)\,dx$ gives the area between the curve $y = f(x)$ The $x$ -axis, and the vertical lines $x = a$ and $x = b$ .

If $f(x)$ changes sign on $[a, b]$ The integral gives the net (signed) area. The total area is Computed by splitting at the zeros of $f$ and taking absolute values:

\mathrm\{Total area\} = \int_a^b |f(x)|\,dx

Adjust the sliders to change the function and limits, and observe how the shaded area approximates the definite integral.

Expand

Evaluate $\int_1^3 (x^2 + 1)\,dx$

\begin`\{aligned}` \int_1^3 (x^2 + 1)\,dx &= \left[\frac\{x^3\}\{3\} + x\right]_1^3 \\[6pt] &= \left(\frac\{27\}\{3\} + 3\right) - \left(\frac\{1\}\{3\} + 1\right) \\[6pt] &= 12 - \frac\{4\}\{3\} = \frac\{32\}\{3\} \end`\{aligned}`

Find the area enclosed by $y = x^2 - 4$ and the $x$ -axis.

The curve crosses the $x$ -axis at $x = -2$ and $x = 2$ .

\begin`\{aligned}` \mathrm\{Area\} &= \int_\{-2\}^2 |x^2 - 4|\,dx = \int_\{-2\}^2 (4 - x^2)\,dx \\[6pt] &= \left[4x - \frac\{x^3\}\{3\}\right]_\{-2\}^2 \\[6pt] &= \left(8 - \frac\{8\}\{3\}\right) - \left(-8 + \frac\{8\}\{3\}\right) = \frac\{32\}\{3\} \end`\{aligned}`

3. Integration Techniques

3.1 Substitution ( $u$ -Substitution)

Substitution is the reverse of the chain rule. Given an integral containing a composite function, Choose a substitution $u = g(x)$ such that $du = g'(x)\,dx$ transforms the integral into a simpler Form.

General procedure:

Identify an inner function and set $u = g(x)$ .
Compute $du = g'(x)\,dx$ and solve for $dx$ .
Rewrite the entire integral in terms of $u$ .
Evaluate the integral with respect to $u$ .
Substitute back $x$ (for indefinite integrals) or change the limits (for definite integrals).

Changing limits for definite integrals:

\int_a^b f(g(x))\,g'(x)\,dx = \int_\{g(a)\}^\{g(b)\} f(u)\,du

Expand

Find $\int 2x\sqrt{x^2 + 1}\,dx$

Let $u = x^2 + 1$ Then $du = 2x\,dx$ .

\int 2x\sqrt\{x^2 + 1\}\,dx = \int \sqrt\{u\}\,du = \frac\{2\}\{3\}u^\{3/2\} + C = \frac\{2\}\{3\}(x^2 + 1)^\{3/2\} + C

Evaluate $\int_0^2 \frac{x}{x^2 + 1}\,dx$

Let $u = x^2 + 1$$du = 2x\,dx$ . When $x = 0$$u = 1$ ; when $x = 2$$u = 5$ .

\int_0^2 \frac\{x\}\{x^2 + 1\}\,dx = \frac\{1\}\{2\}\int_1^5 \frac\{1\}\{u\}\,du = \frac\{1\}\{2\}\Big[\ln|u|\Big]_1^5 = \frac\{1\}\{2\}\ln 5

3.2 Integration by Parts

Integration by parts is the reverse of the product rule. The formula is:

\int u\,dv = uv - \int v\,du

Choosing $u$ and $dv$ : Use the LIATE priority rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ as the function that appears first in the list.

Integration by parts may need to be applied repeatedly. For integrals of the form $\int e^{ax}\sin(bx)\,dx$ or $\int e^{ax}\cos(bx)\,dx$ Apply integration by parts twice and solve The resulting equation algebraically.

Tabular (DI) method: For integrals of the form $\int f(x) g(x)\,dx$ where one factor Differentiates to zero after finitely many steps, the tabular method provides an efficient Alternative to repeated application of the formula.

Expand

Find $\int x e^x\,dx$

Let $u = x$$dv = e^x\,dx$ . Then $du = dx$ and $v = e^x$ .

\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = (x - 1)e^x + C

Find $\int x^2 \sin x\,dx$

Apply integration by parts twice.

First: $u = x^2$$dv = \sin x\,dx$ So $du = 2x\,dx$$v = -\cos x$ .

\int x^2 \sin x\,dx = -x^2 \cos x + 2\int x \cos x\,dx

Second: $u = x$$dv = \cos x\,dx$ So $du = dx$$v = \sin x$ .

\begin`\{aligned}` \int x^2 \sin x\,dx &= -x^2 \cos x + 2\left(x\sin x - \int \sin x\,dx\right) \\[6pt] &= -x^2 \cos x + 2x\sin x + 2\cos x + C \end`\{aligned}`

3.3 Partial Fractions

When the integrand is a rational function $\dfrac{P(x)}{Q(x)}$ where $\deg P < \deg Q$ and $Q$ Factors into linear or irreducible quadratic factors, partial fraction decomposition converts the Integrand into a sum of simpler fractions.

Decomposition rules:

Factor in $Q(x)$	Term in decomposition
$(ax + b)$	$\dfrac{A}{ax + b}$
$(ax + b)^k$	$\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax+b)^2} + \cdots + \dfrac{A_k}{(ax+b)^k}$
$ax^2 + bx + c$ (irreducible)	$\dfrac{Ax + B}{ax^2 + bx + c}$

Expand

Find $\int \frac{5x + 1}{(x+1)(x-2)}\,dx$

Decompose: $\dfrac{5x + 1}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$

5x + 1 = A(x - 2) + B(x + 1)

Setting $x = -1$ : $-5 + 1 = A(-3) \implies A = \frac{4}{3}$

Setting $x = 2$ : $10 + 1 = B(3) \implies B = \frac{11}{3}$

\int \frac\{5x + 1\}\{(x+1)(x-2)\}\,dx = \frac\{4\}\{3\}\ln|x+1| + \frac\{11\}\{3\}\ln|x-2| + C

3.4 Trigonometric Integrals

Several standard techniques apply to integrals involving trigonometric functions.

Pythagorean identities for conversion:

\sin^2 x = \frac\{1 - \cos 2x\}\{2\}, \qquad \cos^2 x = \frac\{1 + \cos 2x\}\{2\}

These are used to reduce the power of trigonometric functions in integrands.

Products to sums:

\sin A \cos B = \frac\{1\}\{2\}[\sin(A+B) + \sin(A-B)]

\cos A \cos B = \frac\{1\}\{2\}[\cos(A-B) + \cos(A+B)]

\sin A \sin B = \frac\{1\}\{2\}[\cos(A-B) - \cos(A+B)]

Using substitution with trigonometric derivatives: Recall that $\dfrac{d}{dx}(\sin x) = \cos x$ And $\dfrac{d}{dx}(\cos x) = -\sin x$ . Many trigonometric integrals yield to $u$ -substitution when the derivative of one trigonometric factor is present.

Integrals of the form $\int \tan^n x\,dx$ : Rewrite in terms of $\sec^2 x$ and use $\dfrac{d}{dx}(\tan x) = \sec^2 x$ .

Expand

Find $\int \sin^2 x\,dx$

\int \sin^2 x\,dx = \int \frac\{1 - \cos 2x\}\{2\}\,dx = \frac\{x\}\{2\} - \frac\{\sin 2x\}\{4\} + C

Find $\int \sin^3 x\,dx$

\begin`\{aligned}` \int \sin^3 x\,dx &= \int \sin x (1 - \cos^2 x)\,dx \\[6pt] &= \int \sin x\,dx - \int \sin x \cos^2 x\,dx \\[6pt] &= -\cos x + \frac\{\cos^3 x\}\{3\} + C \end`\{aligned}`

(using $u = \cos x$$du = -\sin x\,dx$ for the second integral)

4. Applications of Integration

4.1 Area Between Two Curves

Given two continuous functions $f$ and $g$ with $f(x) \geq g(x)$ on $[a, b]$ The area between the Curves is:

A = \int_a^b [f(x) - g(x)]\,dx

If the curves intersect, find the points of intersection and split the integral accordingly so that the integrand $|f(x) - g(x)|$ is always non-negative.

Horizontal strips: When integrating with respect to $y$ The formula becomes:

A = \int_c^d [f(y) - g(y)]\,dy

Where the curves are expressed as $x = f(y)$ and $x = g(y)$ .

Expand

Find the area between $y = x^2$ and $y = 2x$ .

Intersection: $x^2 = 2x \implies x = 0$ or $x = 2$ . On $[0, 2]$$2x \geq x^2$ .

A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac\{x^3\}\{3\}\right]_0^2 = 4 - \frac\{8\}\{3\} = \frac\{4\}\{3\}

4.2 Volumes of Revolution

When a region bounded by $y = f(x)$ The $x$ -axis, and the lines $x = a$$x = b$ is revolved about The $x$ -axis, the volume of the solid of revolution is:

V = \pi \int_a^b [f(x)]^2\,dx

Revolution about the $y$ -axis:

V = \pi \int_c^d [g(y)]^2\,dy

Where $x = g(y)$ .

Revolution about a horizontal line $y = k$ :

V = \pi \int_a^b [f(x) - k]^2\,dx

Washer method (volume between two surfaces):

V = \pi \int_a^b \left([f(x)]^2 - [g(x)]^2\right)\,dx

Where $f(x) \geq g(x) \geq 0$ on $[a, b]$ .

Expand

Find the volume generated by revolving $y = \sqrt{x}$ about the $x$ -axis from $x = 0$ to $x = 4$ .

\begin`\{aligned}` V &= \pi \int_0^4 (\sqrt\{x\})^2\,dx = \pi \int_0^4 x\,dx \\[6pt] &= \pi \left[\frac\{x^2\}\{2\}\right]_0^4 = 8\pi \end`\{aligned}`

4.3 Kinematics

Integration connects the kinematic quantities:

V(t) = \int a(t)\,dt + v_0

S(t) = \int v(t)\,dt + s_0

Where $a(t)$ is acceleration, $v(t)$ is velocity, $s(t)$ is displacement, and $v_0$$s_0$ are Initial conditions.

Key relationships:

The displacement over a time interval $[t_1, t_2]$ is $\displaystyle\int_{t_1}^{t_2} v(t)\,dt$ .
The total distance travelled is $\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$ .
The velocity is the derivative of displacement: $v = \dfrac{ds}{dt}$ .
The acceleration is the derivative of velocity: $a = \dfrac{dv}{dt}$ .

Expand

A particle moves with velocity $v(t) = 6t^2 - 4t$ m/s for $t \geq 0$ . Find the displacement and Total distance in the first 3 seconds.

Displacement:

\int_0^3 (6t^2 - 4t)\,dt = \left[2t^3 - 2t^2\right]_0^3 = 54 - 18 = 36 \mathrm\{ m\}

Total distance: find when $v(t) = 0$ : $6t^2 - 4t = 0 \implies t = 0$ or $t = \frac{2}{3}$ .

\begin`\{aligned}` \mathrm\{Distance\} &= \int_0^\{2/3\} |6t^2 - 4t|\,dt + \int_\{2/3\}^3 |6t^2 - 4t|\,dt \\[6pt] &= \int_0^\{2/3\} (4t - 6t^2)\,dt + \int_\{2/3\}^3 (6t^2 - 4t)\,dt \\[6pt] &= \left[2t^2 - 2t^3\right]_0^\{2/3\} + \left[2t^3 - 2t^2\right]_\{2/3\}^3 \\[6pt] &= \frac\{8\}\{27\} + \left(36 - \left(-\frac\{8\}\{27\}\right)\right) = \frac\{8\}\{27\} + 36 + \frac\{8\}\{27\} = \frac\{988\}\{27\} \approx 36.6 \mathrm\{ m\} \end`\{aligned}`

5. Properties of Definite Integrals

These properties follow from the definition of the definite integral and are essential for Simplifying computations.

Linearity

\int_a^b [k f(x) + l g(x)]\,dx = k\int_a^b f(x)\,dx + l\int_a^b g(x)\,dx

For constants $k, l \in \mathbb{R}$ .

Reversal of Limits

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx

Interval of Zero Length

\int_a^a f(x)\,dx = 0

Additivity Over Intervals

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

For any $c \in [a, b]$ .

Comparison Properties

If $f(x) \geq 0$ for all $x \in [a, b]$ Then $\displaystyle\int_a^b f(x)\,dx \geq 0$ .

If $f(x) \geq g(x)$ for all $x \in [a, b]$ Then $\displaystyle\int_a^b f(x)\,dx \geq \int_a^b g(x)\,dx$ .

Bounds on the Integral

If $m \leq f(x) \leq M$ for all $x \in [a, b]$ Then:

M(b - a) \leq \int_a^b f(x)\,dx \leq M(b - a)

Expand

Given $\int_1^3 f(x)\,dx = 5$ and $\int_1^3 g(x)\,dx = -2$ Find $\int_3^1 [3f(x) - 2g(x)]\,dx$ .

\begin`\{aligned}` \int_3^1 [3f(x) - 2g(x)]\,dx &= -\int_1^3 [3f(x) - 2g(x)]\,dx \\[6pt] &= -\left(3 \cdot 5 - 2 \cdot (-2)\right) = -(15 + 4) = -19 \end`\{aligned}`

6. Improper Integrals (HL)

Improper integrals extend the concept of definite integration to cases where the interval of Integration is unbounded or the integrand has a vertical asymptote within (or at an endpoint of) the Interval of integration.

Type 1: Infinite Limits of Integration

\int_a^\{\infty\} f(x)\,dx = \lim_\{t \to \infty\} \int_a^t f(x)\,dx

\int_\{-\infty\}^b f(x)\,dx = \lim_\{t \to -\infty\} \int_t^b f(x)\,dx

\int_\{-\infty\}^\{\infty\} f(x)\,dx = \int_\{-\infty\}^c f(x)\,dx + \int_c^\{\infty\} f(x)\,dx

Both limits must converge independently for the integral to converge.

Type 2: Integrands with Vertical Asymptotes

If $f$ has a vertical asymptote at $x = a$ :

\int_a^b f(x)\,dx = \lim_\{t \to a^+\} \int_t^b f(x)\,dx

If $f$ has a vertical asymptote at $x = b$ :

\int_a^b f(x)\,dx = \lim_\{t \to b^-\} \int_a^t f(x)\,dx

If $f$ has a vertical asymptote at an interior point $x = c \in (a, b)$ :

\int_a^b f(x)\,dx = \lim_\{t \to c^-\} \int_a^t f(x)\,dx + \lim_\{t \to c^+\} \int_t^b f(x)\,dx

Convergence and Divergence

An improper integral converges if the corresponding limit exists and is finite. It diverges Otherwise.

$p$ -test for convergence: The integral $\displaystyle\int_1^{\infty} \frac{1}{x^p}\,dx$ Converges if and only if $p > 1$ In which case:

\int_1^\{\infty\} \frac\{1\}\{x^p\}\,dx = \frac\{1\}\{p - 1\}

Similarly, $\displaystyle\int_0^1 \frac{1}{x^p}\,dx$ converges if and only if $p < 1$ .

Comparison test: If $0 \leq f(x) \leq g(x)$ for all $x \geq a$ And $\int_a^{\infty} g(x)\,dx$ Converges, then $\int_a^{\infty} f(x)\,dx$ also converges. Conversely, if $\int_a^{\infty} f(x)\,dx$ Diverges, then $\int_a^{\infty} g(x)\,dx$ also diverges.

Expand

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$ converges.

\begin`\{aligned}` \int_1^\{\infty\} \frac\{1\}\{x^2\}\,dx &= \lim_\{t \to \infty\} \int_1^t x^\{-2\}\,dx = \lim_\{t \to \infty\} \left[-\frac\{1\}\{x\}\right]_1^t \\[6pt] &= \lim_\{t \to \infty\} \left(-\frac\{1\}\{t\} + 1\right) = 1 \end`\{aligned}`

The integral converges to $1$ . (Consistent with the $p$ -test: $p = 2 > 1$ .)

Determine whether $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges.

\begin`\{aligned}` \int_0^1 \frac\{1\}\{\sqrt\{x\}\}\,dx &= \lim_\{t \to 0^+\} \int_t^1 x^\{-1/2\}\,dx = \lim_\{t \to 0^+\} \Big[2\sqrt\{x\}\Big]_t^1 \\[6pt] &= \lim_\{t \to 0^+\} (2 - 2\sqrt\{t\}) = 2 \end`\{aligned}`

The integral converges to $2$ . (Consistent with the $p$ -test: $p = \frac{1}{2} < 1$ .)

7. Practice Problems

Find

\int \frac{x^3 + 2x}{\sqrt{x^2 + 1}}\,dx

Let

u = x^2 + 1$$du = 2x\,dx

. Note that

x^3 + 2x = x(x^2 + 1) + x = xu + \frac{u - 1}{2} + \frac{1}{2}

Alternatively, split: $\dfrac{x^3 + 2x}{\sqrt{x^2 + 1}} = \dfrac{x(x^2 + 1)}{\sqrt{x^2 + 1}} + \dfrac{x}{\sqrt{x^2 + 1}} = x\sqrt{x^2 + 1} + \dfrac{x}{\sqrt{x^2 + 1}}$ .

\begin`\{aligned}` \int x\sqrt\{x^2 + 1\}\,dx &= \frac\{1\}\{3\}(x^2 + 1)^\{3/2\} + C_1 \\[6pt] \int \frac\{x\}\{\sqrt\{x^2 + 1\}\}\,dx &= \sqrt\{x^2 + 1\} + C_2 \end`\{aligned}`

\int \frac\{x^3 + 2x\}\{\sqrt\{x^2 + 1\}\}\,dx = \frac\{1\}\{3\}(x^2 + 1)^\{3/2\} + \sqrt\{x^2 + 1\} + C

Find the volume of revolution when the region bounded by

y = \ln x$$x = e

And the

x

-axis is rotated about the

x

-axis.

The curve

y = \ln x

meets the

x

-axis at

x = 1

\begin`\{aligned}` V &= \pi \int_1^e (\ln x)^2\,dx \end`\{aligned}`

Integration by parts with $u = (\ln x)^2$$dv = dx$ : $du = \dfrac{2\ln x}{x}\,dx$$v = x$ .

\begin`\{aligned}` \int (\ln x)^2\,dx &= x(\ln x)^2 - 2\int \ln x\,dx \end`\{aligned}`

Apply parts again for $\int \ln x\,dx$ : $u = \ln x$$dv = dx$ So $\int \ln x\,dx = x\ln x - x$ .

\begin`\{aligned}` \int (\ln x)^2\,dx &= x(\ln x)^2 - 2(x\ln x - x) + C \\[6pt] &= x(\ln x)^2 - 2x\ln x + 2x + C \end`\{aligned}`

Evaluating from $1$ to $e$ :

\begin`\{aligned}` \Big[x(\ln x)^2 - 2x\ln x + 2x\Big]_1^e &= (e - 2e + 2e) - (0 - 0 + 2) = e - 2 \end`\{aligned}`

V = \pi(e - 2)

Use partial fractions to find

\int \frac{3x + 7}{(x + 2)(x - 1)}\,dx

\frac\{3x + 7\}\{(x+2)(x-1)\} = \frac\{A\}\{x+2\} + \frac\{B\}\{x-1\}

3x + 7 = A(x - 1) + B(x + 2)

$x = 1$ : $10 = 3B \implies B = \dfrac{10}{3}$

$x = -2$ : $1 = -3A \implies A = -\dfrac{1}{3}$

\int \frac\{3x + 7\}\{(x+2)(x-1)\}\,dx = -\frac\{1\}\{3\}\ln|x+2| + \frac\{10\}\{3\}\ln|x-1| + C

Evaluate

\int \frac{x^2}{x^2 - 1}\,dx

using polynomial long division and partial fractions.

Since

\deg(\mathrm{numerator}) = \deg(\mathrm{denominator})

Perform long division first:

\frac\{x^2\}\{x^2 - 1\} = 1 + \frac\{1\}\{x^2 - 1\} = 1 + \frac\{1\}\{(x-1)(x+1)\}

Partial fractions: $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$

$1 = A(x+1) + B(x-1)$

$x = 1$ : $A = \dfrac{1}{2}$$x = -1$ : $B = -\dfrac{1}{2}$

\begin`\{aligned}` \int \frac\{x^2\}\{x^2 - 1\}\,dx &= \int \left(1 + \frac\{1\}\{2(x-1)\} - \frac\{1\}\{2(x+1)\}\right)\,dx \\[6pt] &= x + \frac\{1\}\{2\}\ln|x-1| - \frac\{1\}\{2\}\ln|x+1| + C \end`\{aligned}`

Find

\int e^x \cos x\,dx

using integration by parts twice.

Let

I = \displaystyle\int e^x \cos x\,dx

First application: $u = e^x$$dv = \cos x\,dx$ So $du = e^x\,dx$$v = \sin x$ .

I = e^x \sin x - \int e^x \sin x\,dx

Second application on $\int e^x \sin x\,dx$ : $u = e^x$$dv = \sin x\,dx$ So $du = e^x\,dx$ $v = -\cos x$ .

\int e^x \sin x\,dx = -e^x \cos x + \int e^x \cos x\,dx = -e^x \cos x + I

Substituting back:

I = e^x \sin x - (-e^x \cos x + I) = e^x \sin x + e^x \cos x - I

2I = e^x (\sin x + \cos x) \implies I = \frac\{e^x (\sin x + \cos x)\}\{2\} + C

Cross-References

Differentiation — Integration is the inverse operation of differentiation. See the Number and Algebra notes for function fundamentals including derivative rules that motivate integration techniques.
Functions — Domain and range considerations determine when antiderivatives are valid. See Number and Algebra for function fundamentals.
Complex Numbers — The exponential form $e^{i\theta} = \cos\theta + i\sin\theta$ provides an elegant derivation of trigonometric integral results. See Complex Numbers.
Logic — Proof techniques (direct …/1-number-and-algebra/3_proof-and-logic, contradiction) are used to justify properties of integrals. See Logic.

Topic	Site	Link
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[Integration]	IB	View

Intuition

Integration is accumulation. When you integrate a velocity function, you are adding up infinitely many tiny displacements to find total distance. The definite integral measures the net area between a curve and the axis, positive above, negative below. The Fundamental Theorem of Calculus reveals that differentiation and integration are inverse operations, like locking and unlocking a door. Substitution is the chain rule backwards, integration by parts is the product rule backwards, and partial fractions拆 apart complex fractions into simple building blocks. Every integration technique is really a strategy for making an unfamiliar integral look like something you already know how to do.

Common Pitfalls

Forgetting the chain rule when integrating composite functions. Look for an inner function and its derivative.
Confusing definite and indefinite integrals. Definite integrals give a numerical value; indefinite integrals give a family of functions.
Incorrectly applying limits of integration after a substitution. Remember to change the limits to the new variable.
Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).
Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.
Cancelling terms instead of factors, $\frac{ab + ac}{a}$ simplifies to $b + c$ , not $bc$ .

Summary

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

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