How to Solve Integrals: Step-by-Step Methods

An integral is, intuitively, an accumulation. The function $f(x)$ tells you how much of something is happening at each point. The integral

${\int }_a^{b}f(x)dx$

tells you the total amount between $x=a$ and $x=b$. If $f(x)$ represents speed at time $x$, the integral is total distance traveled. If $f(x)$ represents the cross-sectional area of a solid at position $x$, the integral is the volume. If $f(x)$ represents the probability density of a random variable, the integral is the probability of falling in the interval $[a,b]$.

This article covers what integration actually is, the fundamental theorem of calculus that makes integration computable, the basic integration rules, the substitution technique (the most-used method), integration by parts, trigonometric integrals and substitutions, partial fractions, improper integrals, numerical methods for non-integrable functions, common pitfalls, and a structured approach for choosing the right technique.

What integration actually is

Two different intuitions converge on the same operation.

Intuition 1 — Area under a curve.

Draw the graph of a positive function $f(x)$. The integral ${\int }_a^{b}f(x)dx$ is the area of the region bounded above by the curve, below by the x-axis, and on the sides by the vertical lines $x=a$ and $x=b$.

This is the geometric interpretation, and it is how integration was originally understood. Archimedes used a precursor of integration (the method of exhaustion) to compute areas of curved regions in the third century BCE.

Intuition 2 — Accumulation.

If $f(t)$ describes a rate of change — speed, water flow, marginal cost — then ${\int }_a^{b}f(t)dt$ is the total amount accumulated between $t=a$ and $t=b$.

This is the physical interpretation, and it is the most useful for applications. A car going at constantly changing speed travels a distance equal to the integral of its speed.

The Riemann sum makes both intuitions precise. Divide the interval $[a,b]$ into $n$ small subintervals of width $\Delta x=\frac{b-a}{n}$. In each subinterval, pick a point $x_i^{\ast }$. The sum

${\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x$

approximates the area (or the accumulation). As $n\to \infty$ (the subintervals get smaller), the sum approaches the exact integral:

${\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x$

This is the formal definition. In practice, you almost never compute integrals from the definition — you use the techniques described below.

The fundamental theorem of calculus

The fundamental theorem says: integration and differentiation are inverse operations.

Formally, if $F$ is any antiderivative of $f$ (meaning $F^{\prime }(x)=f(x)$), then

${\int }_a^{b}f(x)dx=F(b)-F(a)$

This is enormous. It says that to compute an integral, you do not need to compute Riemann sums. You just need to find an antiderivative of the integrand, and evaluate it at the endpoints.

Example 1: Compute ${\int }_0^{2}3x^{2}dx$.

The antiderivative of $3x^2$ is $x^3$ (you can verify: $\frac{d}{dx}(x^3)=3x^2$). So:

${\int }_0^{2}3x^{2}dx=[x^3{]}_0^2=2^3-0^3=8$

The whole integration boils down to finding an antiderivative. The techniques below are techniques for finding antiderivatives.

Basic integration rules

Memorize these. They are the building blocks.

Power rule (for integration):

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\text{for }n\ne -1$

The constant $C$ is the constant of integration: any constant differentiates to zero, so the antiderivative is determined only up to an added constant. (For definite integrals — those with limits like ${\int }_a^b$ — the constants cancel when you evaluate at endpoints.)

For $n=-1$:

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

Exponential and logarithmic:

$\int e^{x}dx=e^x+C$ $\int a^{x}dx=\frac{a^x}{\ln a}+C$

Trigonometric:

$\int \sin xdx=-\cos x+C$ $\int \cos xdx=\sin x+C$ $\int {\sec }^{2}xdx=\tan x+C$

Linearity:

$\int (af(x)+bg(x))dx=a\int f(x)dx+b\int g(x)dx$

This means you can split sums and pull constants outside integrals.

Example 2: Compute $\int (3x^2-4x+5)dx$.

Apply linearity, then the power rule:

$\int 3x^{2}dx-\int 4xdx+\int 5dx=x^3-2x^2+5x+C$

Substitution (u-substitution): the most-used technique

When the integrand contains a function and (something close to) its derivative, substitution simplifies the integral.

The setup. If the integrand is of the form $f(g(x))\cdot g^{\prime }(x)$, then setting $u=g(x)$ gives $du=g^{\prime }(x)dx$, and the integral becomes

$\int f(g(x))\cdot g^{\prime }(x)dx=\int f(u)du$

The original integral, which might have been hard, becomes a simpler integral in $u$.

Example 3: Compute $\int 2x\cdot e^{x^2}dx$.

Notice that $\frac{d}{dx}(x^2)=2x$. Set $u=x^2$, so $du=2xdx$. The integral becomes:

$\int e^{u}du=e^u+C=e^{x^2}+C$

Example 4: Compute $\int \frac{2x}{1+x^2}dx$.

Set $u=1+x^2$, so $du=2xdx$. The integral becomes:

$\int \frac{1}{u}du=\ln |u|+C=\ln (1+x^2)+C$

(We can drop the absolute value because $1+x^2>0$ always.)

Example 5 — substitution that requires adjustment: Compute $\int x\sqrt{x^2+1}dx$.

Set $u=x^2+1$, so $du=2xdx$, which means $xdx=\frac{1}{2}du$. The integral becomes:

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

The key recognition skill: looking at an integral and spotting "this is a function of something whose derivative appears." Practice this with many examples.

Integration by parts

When substitution does not work, integration by parts often does. It comes from the product rule for differentiation.

Formula:

$\int udv=uv-\int vdu$

To use it: identify part of the integrand to call $u$, the rest to call $dv$. Differentiate $u$ to get $du$; integrate $dv$ to get $v$. Substitute.

The choice of $u$ and $dv$ matters. A common mnemonic: LIATE — choose $u$ in this order of preference:

• Logarithmic functions
• Inverse trig functions
• Algebraic functions (polynomials)
• Trigonometric functions
• Exponential functions

Example 6: Compute $\int x{e}^{x}dx$.

Algebraic comes before exponential in LIATE, so set $u=x$, $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$. Apply the formula:

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=e^x(x-1)+C$

Example 7: Compute $\int x\cos xdx$.

Set $u=x$, $dv=\cos xdx$. Then $du=dx$, $v=\sin x$. Apply:

$\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C$

Example 8 — integration by parts twice: Compute $\int x^2{e}^{x}dx$.

Set $u=x^2$, $dv=e^{x}dx$. Then $du=2xdx$, $v=e^x$. Apply:

$\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx=x^2{e}^x-2\int x{e}^{x}dx$

We computed $\int x{e}^{x}dx=e^x(x-1)+C$ above. Substitute:

$\int x^2{e}^{x}dx=x^2{e}^x-2e^x(x-1)+C=e^x(x^2-2x+2)+C$

Trigonometric integrals and substitutions

A small library of techniques handles most integrals involving trigonometric functions.

Powers of sine and cosine. Use the identities ${\sin }^{2}x=\frac{1-\cos (2x)}{2}$ and ${\cos }^{2}x=\frac{1+\cos (2x)}{2}$ for even powers, or save one factor and convert the rest for odd powers.

Example 9: Compute $\int {\sin }^{3}xdx$.

Rewrite: ${\sin }^{3}x=\sin x\cdot (1-{\cos }^{2}x)$. Substitute $u=\cos x$, $du=-\sin xdx$:

$\int \sin x(1-{\cos }^{2}x)dx=-\int (1-u^2)du=-u+\frac{u^3}{3}+C=-\cos x+\frac{{\cos }^{3}x}{3}+C$

Trigonometric substitution. When an integral contains $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$, substituting a trigonometric function for $x$ often simplifies it.

For $\sqrt{a^2-x^2}$: set $x=a\sin \theta$. For $\sqrt{a^2+x^2}$: set $x=a\tan \theta$. For $\sqrt{x^2-a^2}$: set $x=a\sec \theta$.

Example 10: Compute $\int \sqrt{4-x^2}dx$.

Set $x=2\sin \theta$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$. The integral becomes:

$\int 2\cos \theta \cdot 2\cos \theta d\theta =4\int {\cos }^2\theta d\theta =4\int \frac{1+\cos (2\theta )}{2}d\theta =2\theta +\sin (2\theta )+C$

Substitute back. Since $x=2\sin \theta$, we have $\theta =\arcsin (x/2)$ and $\sin (2\theta )=2\sin \theta \cos \theta =2\cdot \frac{x}{2}\cdot \frac{\sqrt{4-x^2}}{2}=\frac{x\sqrt{4-x^2}}{2}$.

$\int \sqrt{4-x^2}dx=2\arcsin \left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2}+C$

Partial fractions

When the integrand is a rational function $\frac{P(x)}{Q(x)}$ where the degree of $P$ is less than the degree of $Q$, factor $Q(x)$ and decompose the fraction into simpler pieces.

Example 11: Compute $\int \frac{1}{x^2-1}dx$.

Factor: $x^2-1=(x-1)(x+1)$. Decompose:

$\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$

Multiply both sides by $(x-1)(x+1)$: $1=A(x+1)+B(x-1)$. Set $x=1$: $1=2A$, so $A=1/2$. Set $x=-1$: $1=-2B$, so $B=-1/2$.

$\int \frac{1}{x^2-1}dx=\frac{1}{2}\int \frac{1}{x-1}dx-\frac{1}{2}\int \frac{1}{x+1}dx=\frac{1}{2}\ln |x-1|-\frac{1}{2}\ln |x+1|+C=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C$

For repeated roots and quadratic factors, the decomposition is more elaborate, but the principle is the same.

Improper integrals

An improper integral is one where either the interval is infinite, or the integrand is unbounded somewhere in the interval.

Type 1 — infinite interval:

${\int }_1^{\infty }\frac{1}{x^2}dx={\lim }_{b\to \infty }{\int }_1^b\frac{1}{x^2}dx={\lim }_{b\to \infty }{\left[-\frac{1}{x}\right]}_1^b={\lim }_{b\to \infty }\left(1-\frac{1}{b}\right)=1$

The integral converges to 1.

Type 2 — unbounded integrand:

${\int }_0^1\frac{1}{\sqrt{x}}dx={\lim }_{a\to 0^{+}}{\int }_a^1\frac{1}{\sqrt{x}}dx={\lim }_{a\to 0^{+}}[2\sqrt{x}{]}_a^1={\lim }_{a\to 0^{+}}(2-2\sqrt{a})=2$

Converges to 2.

Some improper integrals diverge — they fail to be finite. The integral ${\int }_1^{\infty }\frac{1}{x}dx$ diverges (the antiderivative $\ln x$ grows without bound).

Numerical integration

Many integrals cannot be expressed in closed form. The integral ${\int }_0^x{e}^{-t^2}dt$ (central to probability theory — it gives the area under the normal distribution up to $x$) has no antiderivative in terms of elementary functions.

When closed form fails, numerical methods compute approximate values:

Trapezoidal rule:

${\int }_a^{b}f(x)dx\approx \frac{\Delta x}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots +2f(x_{n-1})+f(x_n)\right]$

where $\Delta x=(b-a)/n$ and $x_i=a+i\Delta x$.

Simpson's rule is more accurate:

${\int }_a^{b}f(x)dx\approx \frac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\cdots +4f(x_{n-1})+f(x_n)\right]$

(requires $n$ to be even).

For software computation, libraries like SciPy's quad use adaptive methods that achieve high accuracy automatically.

Common pitfalls

Pitfall 1 — Forgetting the constant of integration. For indefinite integrals (without limits), always include the $+C$. Forgetting it is a standard exam mistake.

Pitfall 2 — Sign errors in integration by parts. The formula is $\int udv=uv-\int vdu$. The minus sign in front of the second integral is missed regularly. Double-check.

Pitfall 3 — Forgetting to change limits when using substitution in definite integrals. When you set $u=g(x)$, the original limits $a$ and $b$ become $g(a)$ and $g(b)$. If you forget this, your answer will be wrong.

Example 12: Compute ${\int }_0^{1}2x{e}^{x^2}dx$.

Substitute $u=x^2$, $du=2xdx$. When $x=0$, $u=0$; when $x=1$, $u=1$. So:

${\int }_0^1{e}^{u}du=e^1-e^0=e-1$

(About 1.718.)

Pitfall 4 — Trying the wrong technique. Most integrals can be solved by one of substitution, parts, or partial fractions. If you have spent ten minutes on a problem and made no progress, the technique is wrong. Try another.

A structured approach to choosing the technique

When you encounter an unfamiliar integral, work through this checklist:

1. Is it a basic form? Check the table of basic integrals first.
2. Does substitution apply? Look for $f(g(x))\cdot g^{\prime }(x)$ patterns. This is the most common technique.
3. Is the integrand a product? Try integration by parts. LIATE for choice of $u$.
4. Does it involve $\sqrt{a^2\pm x^2}$ or $\sqrt{x^2-a^2}$? Try trigonometric substitution.
5. Is it a rational function? Try partial fractions.
6. Does the integrand have trig powers? Use trig identities or save-one-factor techniques.
7. None of the above? It might be a non-elementary integral. Try numerical methods.

Most students can solve 90% of textbook integrals with substitution, by parts, and partial fractions. The remaining 10% require trigonometric substitution or numerical methods.