Part 5 · Chapter 25

Indefinite Integration

Running the derivative backwards — recovering a function from its rate of change, building the table of standard integrals, and the handful of techniques that turn almost any expression into something you can integrate

Fundamentals of Mathematics Prof. Mithun Mondal Reading time ≈ 48 min
i What you'll learn
  • The antiderivative as the reverse of the derivative, and why every integral carries a constant of integration \(+C\).
  • The table of standard integrals read straight off the derivative rules, plus the linearity of integration.
  • Integration by substitution, which reverses the chain rule.
  • Integration by parts, which reverses the product rule, and the LIATE guide for choosing parts.
  • Partial fractions for rational functions and the reduction of trigonometric integrands.
  • The standard special forms — recognised by completing the square.
Section 25-1

The Antiderivative

Differentiation takes a function and reports its slope. Integration asks the opposite question: given the slope \(f(x)\) at every point, which function \(F\) has it? Such an \(F\) is an antiderivative of \(f\), written \(F'(x)=f(x)\). Because the derivative of a constant is zero, adding any constant to \(F\) leaves its slope unchanged — so antiderivatives come not one at a time but as a whole family of vertically shifted curves.

+C +C same slope
Every curve in the family has the same \(f(x)\): they differ only by the constant \(C\)
∫
The indefinite integral
\(\displaystyle\int f(x)\,dx=F(x)+C,\qquad\text{where } F'(x)=f(x)\)

The symbol \(\int\) is an elongated "S" (for sum); \(dx\) names the variable of integration. The result is a family of functions, and the constant of integration \(C\) is not optional — dropping it is the single most common error. Integration and differentiation undo each other: \(\tfrac{d}{dx}\!\int f\,dx=f\).

Section 25-2

Standard Integrals

Every derivative rule, read from right to left, becomes an integration formula. Memorising this short table is what makes the rest of the chapter possible — most problems are an exercise in reshaping an integrand until it matches one of these entries.

F(x) f(x) d/dx ∫ … dx
Each line of the table is a derivative rule run backwards
Table of standard integrals
\[ \begin{array}{ll} \displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\ (n\neq-1) & \displaystyle\int \frac{1}{x}\,dx=\ln|x|+C \\[1.1em] \displaystyle\int e^{x}\,dx=e^{x}+C & \displaystyle\int a^{x}\,dx=\frac{a^{x}}{\ln a}+C \\[1.1em] \displaystyle\int \sin x\,dx=-\cos x+C & \displaystyle\int \cos x\,dx=\sin x+C \\[1.1em] \displaystyle\int \sec^{2}x\,dx=\tan x+C & \displaystyle\int \csc^{2}x\,dx=-\cot x+C \\[1.1em] \displaystyle\int \sec x\tan x\,dx=\sec x+C & \displaystyle\int \csc x\cot x\,dx=-\csc x+C \\[1.1em] \displaystyle\int \frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x+C & \displaystyle\int \frac{dx}{1+x^{2}}=\tan^{-1}x+C \end{array} \]
📋
The power rule and its one exception
\(\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\ (n\neq-1);\qquad \int x^{-1}\,dx=\ln|x|+C\)

Raise the power by one and divide by the new power — except at \(n=-1\), where that recipe would divide by zero and the answer is instead \(\ln|x|\). The absolute value keeps the logarithm defined for negative \(x\) as well.

Section 25-3

Rules & Linearity

Integration inherits two simple rules from differentiation: a constant multiplier slides outside the integral, and the integral of a sum is the sum of the integrals. Together these make integration linear, so any polynomial can be integrated term by term.

➕
Linearity of the integral
\(\displaystyle\int\!\big[a\,f(x)+b\,g(x)\big]\,dx=a\!\int\! f(x)\,dx+b\!\int\! g(x)\,dx\)

Constants pass through the integral sign, and sums split apart. There is, however, no product rule and no quotient rule for integration — \(\int fg\,dx\neq\big(\int f\big)\big(\int g\big)\). That missing shortcut is exactly why the techniques of the next sections exist.

Section 25-4

Integration by Substitution

The chain rule says \(\tfrac{d}{dx}F(g(x))=F'(g(x))\,g'(x)\). Read backwards, it tells us how to integrate any product that hides a function and its own derivative: replace the inner function by a new variable \(u=g(x)\), carry \(du=g'(x)\,dx\) along with it, integrate in the simpler \(u\)-world, then translate back.

x–world ∫f(g)g'dx u–world ∫f(u)du u = g(x) back-substitute
Substitute \(u=g(x)\), integrate in \(u\), then return to \(x\)
🔄
The substitution formula
\(\displaystyle\int f\big(g(x)\big)\,g'(x)\,dx=\int f(u)\,du,\qquad u=g(x)\)

Two patterns appear constantly: \(\displaystyle\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\) and \(\displaystyle\int [f(x)]^{n}f'(x)\,dx=\frac{[f(x)]^{n+1}}{n+1}+C\). The whole method stands or falls on converting \(dx\) into \(du\) — never leave a stray \(dx\) behind.

Section 25-5

Integration by Parts

The product rule \((uv)'=u'v+uv'\), integrated and rearranged, becomes a tool for integrals that are a product of two unlike functions — an \(x\) times an exponential, a logarithm on its own, a polynomial times a sine. You trade the integral you cannot do for one you can.

∫v du ∫u dv u v
The two areas plus the corner rectangle tile \(uv\): \(\int u\,dv+\int v\,du=uv\)
🧩
The integration-by-parts formula
\(\displaystyle\int u\,dv = uv-\int v\,du\)

Split the integrand into a part to differentiate (\(u\)) and a part to integrate (\(dv\)). A good choice makes the new integral \(\int v\,du\) simpler than the one you started with; a bad choice makes it worse. For \(\int\ln x\,dx\), take \(u=\ln x,\ dv=dx\) — the lone logarithm becomes integrable.

! Choosing \(u\): the LIATE order

When the integrand is a product, pick \(u\) as whichever factor comes first in the order Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential — and let the rest be \(dv\). The idea is to differentiate the function that simplifies when differentiated (a logarithm or a polynomial) and integrate the one that stays tame (a sine or an exponential).

Section 25-6

Partial Fractions

A ratio of polynomials looks forbidding until it is broken into a sum of simple pieces, each of which integrates to a logarithm or an arctangent. The method of partial fractions reverses the act of putting fractions over a common denominator — provided the numerator's degree is lower than the denominator's (if not, divide first).

➗
Decomposing a proper rational function
\(\displaystyle\frac{1}{(x-a)(x-b)}=\frac{1}{a-b}\!\left(\frac{1}{x-a}-\frac{1}{x-b}\right)\)

Each distinct linear factor \((x-a)\) contributes a term \(\tfrac{A}{x-a}\); a repeated factor \((x-a)^2\) contributes \(\tfrac{A}{x-a}+\tfrac{B}{(x-a)^2}\); an irreducible quadratic \((x^2+px+q)\) contributes \(\tfrac{Ax+B}{x^2+px+q}\). Find the constants by matching coefficients or substituting convenient values.

Section 25-7

Trigonometric Integrals

Powers and products of sines and cosines are not in the standard table, but trigonometric identities rewrite them until they are. The workhorses are the power-reduction formulas, which trade a squared term for a plain cosine of the double angle.

Four more standard trigonometric integrals
\[ \begin{array}{ll} \displaystyle\int \tan x\,dx=\ln|\sec x|+C & \displaystyle\int \cot x\,dx=\ln|\sin x|+C \\[1.1em] \displaystyle\int \sec x\,dx=\ln|\sec x+\tan x|+C & \displaystyle\int \csc x\,dx=\ln|\csc x-\cot x|+C \end{array} \]
🌊
Power reduction
\(\displaystyle \sin^{2}x=\frac{1-\cos 2x}{2},\qquad \cos^{2}x=\frac{1+\cos 2x}{2}\)

These flatten a squared sine or cosine into something the table handles directly. For odd powers, peel off one factor and convert the rest with \(\sin^2x+\cos^2x=1\), then substitute. For products like \(\sin mx\cos nx\), use the product-to-sum identities first.

Section 25-8

Special Forms

A small gallery of integrals built around \(a^2\pm x^2\) appears so often that they are worth knowing on sight. The trick that unlocks most of them is completing the square: any \(ax^2+bx+c\) in the denominator can be reshaped into one of these forms by a shift of variable.

θ a x √(a²−x²)
With \(x=a\sin\theta\): hypotenuse \(a\), opposite \(x\), adjacent \(\sqrt{a^2-x^2}\)
Standard special forms
\[ \begin{array}{ll} \displaystyle\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C & \displaystyle\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\frac{x}{a}+C \\[1.2em] \displaystyle\int \frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}\ln\!\left|\frac{x-a}{x+a}\right|+C & \displaystyle\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=\ln\!\left|x+\sqrt{x^{2}+a^{2}}\right|+C \\[1.2em] \displaystyle\int \frac{dx}{a^{2}-x^{2}}=\frac{1}{2a}\ln\!\left|\frac{a+x}{a-x}\right|+C & \displaystyle\int \frac{dx}{\sqrt{x^{2}-a^{2}}}=\ln\!\left|x+\sqrt{x^{2}-a^{2}}\right|+C \end{array} \]
√
Recognise, then complete the square
\(\displaystyle ax^{2}+bx+c=a\!\left(x+\tfrac{b}{2a}\right)^{2}+\left(c-\tfrac{b^{2}}{4a}\right)\)

Completing the square turns an awkward quadratic into \((x+k)^2\pm a^2\), and a shift \(t=x+k\) drops it straight onto one of the forms above. The square-root forms can also be cracked by trigonometric substitution — \(x=a\sin\theta\) for \(a^2-x^2\), \(x=a\tan\theta\) for \(a^2+x^2\).

Worked Examples

Putting It to Work

1 Term-by-term with the power rule

Problem. Evaluate \(\displaystyle\int (3x^{2}-4x+5)\,dx\).

Solution. Integrate each term and raise its power by one:

Working
\[ \int (3x^{2}-4x+5)\,dx = x^{3}-2x^{2}+5x+C \]
2 A substitution

Problem. Evaluate \(\displaystyle\int \frac{x}{x^{2}+1}\,dx\).

Solution. Let \(u=x^{2}+1\), so \(du=2x\,dx\) and \(x\,dx=\tfrac12\,du\):

Working
\[ \int \frac{x}{x^{2}+1}\,dx=\frac12\int\frac{du}{u}=\frac12\ln|u|+C=\frac12\ln(x^{2}+1)+C \]
3 Integration by parts

Problem. Evaluate \(\displaystyle\int x\,e^{x}\,dx\).

Solution. By LIATE take \(u=x,\ dv=e^{x}dx\), so \(du=dx,\ v=e^{x}\):

Working
\[ \int x\,e^{x}\,dx=x\,e^{x}-\int e^{x}\,dx=x\,e^{x}-e^{x}+C=(x-1)e^{x}+C \]
4 Partial fractions

Problem. Evaluate \(\displaystyle\int \frac{dx}{(x-1)(x+2)}\).

Solution. Write \(\tfrac{1}{(x-1)(x+2)}=\tfrac{1/3}{x-1}-\tfrac{1/3}{x+2}\), then integrate each piece:

Working
\[ \int \frac{dx}{(x-1)(x+2)}=\frac13\ln|x-1|-\frac13\ln|x+2|+C=\frac13\ln\!\left|\frac{x-1}{x+2}\right|+C \]
5 A trigonometric integral

Problem. Evaluate \(\displaystyle\int \sin^{2}x\,dx\).

Solution. Use power reduction \(\sin^{2}x=\tfrac{1-\cos 2x}{2}\):

Working
\[ \int \sin^{2}x\,dx=\frac12\int(1-\cos 2x)\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C \]
6 A special form

Problem. Evaluate \(\displaystyle\int \frac{dx}{x^{2}+9}\).

Solution. Match \(\int\tfrac{dx}{x^2+a^2}\) with \(a=3\):

Working
\[ \int \frac{dx}{x^{2}+9}=\frac13\tan^{-1}\frac{x}{3}+C \]
Review

Chapter Summary

✓ Antiderivative

\(\int f\,dx=F+C\) with \(F'=f\); never drop the constant \(C\).

✓ Standard integrals

Power rule \(\int x^n\,dx=\tfrac{x^{n+1}}{n+1}\); \(\int\tfrac{dx}{x}=\ln|x|\) at \(n=-1\).

✓ Substitution

Reverses the chain rule: set \(u=g(x)\), convert \(dx\to du\).

✓ By parts

\(\int u\,dv=uv-\int v\,du\); choose \(u\) by LIATE.

✓ Partial fractions

Split a proper rational function into linear and quadratic pieces.

✓ Special forms

Complete the square to land on an \(a^2\pm x^2\) form.

Practice

Problems

Name the method first — standard table, substitution, parts, partial fractions, trig reduction, or special form — then apply the matching tool. Difficulty rises down the list.

  1. Evaluate \(\displaystyle\int (4x^{3}-6x^{2}+2)\,dx\).
  2. Evaluate \(\displaystyle\int \left(\frac{3}{x}+2\cos x\right)dx\).
  3. Evaluate \(\displaystyle\int \left(e^{x}+\sec^{2}x\right)dx\).
  4. Use substitution to find \(\displaystyle\int 2x\,(x^{2}+1)^{5}\,dx\).
  5. Evaluate \(\displaystyle\int \frac{x}{x^{2}+4}\,dx\).
  6. Evaluate \(\displaystyle\int \frac{\sin x}{1+\cos x}\,dx\).
  7. Evaluate \(\displaystyle\int x\sin x\,dx\) by parts.
  8. Evaluate \(\displaystyle\int \ln x\,dx\) (take \(dv=dx\)).
  9. Evaluate \(\displaystyle\int \cos^{2}x\,dx\).
  10. Use partial fractions to find \(\displaystyle\int \frac{dx}{(x-2)(x+1)}\).
  11. Evaluate \(\displaystyle\int \frac{dx}{x^{2}+16}\).
  12. Evaluate \(\displaystyle\int \frac{dx}{\sqrt{25-x^{2}}}\).
  13. Evaluate \(\displaystyle\int x^{2}e^{x}\,dx\) (apply parts twice).
Tip: before reaching for a heavy technique, always check whether the integrand is already in the table, or one substitution away from it. Look for a function sitting beside its own derivative — that signals substitution. A product of two unlike functions signals integration by parts; a ratio of polynomials signals partial fractions; a squared sine or cosine signals power reduction; and an \(a^2\pm x^2\) in the denominator signals a special form, usually after completing the square. And whatever you do, finish by adding the constant \(C\).