MathematicsClass 12ICSE

How Integration by Parts Undoes the Product Rule

Integrate products with integration by parts, split proper rational functions into partial fractions with linear, repeated and irreducible quadratic factors, and integrate improper rational functions after division.

What do you do when an integrand is a product or a fraction?

Substitution works when a function sits beside its derivative, but products like

x e^{x}

and fractions like

\frac{1}{x ^{2} - 1}

need other tools. Integration by parts reverses the product rule, and partial fractions break a complicated fraction into simple pieces.

This lesson covers integration by parts, partial fractions with linear, repeated and quadratic factors, and improper rational functions.

How do you integrate a product of functions using integration by parts?

**Integration by parts states that

\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x

, where u is chosen as the factor that becomes simpler when differentiated — usually following the order inverse trigonometric, logarithmic, algebraic, trigonometric, exponential.

Where it comes from.** Integrate the product rule

(u w)^{'} = u^{'} w + u w^{'}

and rearrange, with

w = \int v d x

.

Worked example. Find

\int x e^{x} d x

with

u = x

and

v = e^{x}

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

Worked example 2. Find

\int lo g x d x

with

u = lo g x

and

v = 1

\int lo g x d x = x lo g x - \int x \times \frac{1}{x} d x = x lo g x - x + C

Worked example 3 (applied twice).

\int x^{2} sin x d x = - x^{2} cos x + 2 \int x cos x d x = - x^{2} cos x + 2 x sin x + 2 cos x + C

.

A special form.

\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C

, so

\int e^{x} (sin x + cos x) d x = e^{x} sin x + C

.

An everyday example. **The total amount of a medicine released into the blood when the release rate follows

t e^{- t}

is found by integration by parts.

The substance. A poor choice of u makes the integral harder** — taking

u = e^{x}

\int x e^{x} d x

produces

\int \frac{x ^{2}}{2} e^{x} d x

, which is worse.

How do you split a proper rational function with linear and repeated linear factors into partial fractions?

**A proper fraction whose denominator has distinct linear factors splits as

\frac{A}{x - a} + \frac{B}{x - b}

, a repeated factor

(x - a)^{2}

needs both

\frac{A}{x - a}

and

\frac{B}{( x - a ) ^{2}}

, and the constants are found by clearing denominators and substituting convenient values of x.

Worked example (distinct factors).**

\frac{x + 5}{( x - 1 ) ( x + 2 )} = \frac{A}{x - 1} + \frac{B}{x + 2}

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

. Putting

x = 1

gives

A = 2

, and putting

x = - 2

gives

B = - 1

. Hence

\int \frac{x + 5}{( x - 1 ) ( x + 2 )} d x = 2 lo g ∣ x - 1∣ - lo g ∣ x + 2∣ + C

Worked example (repeated factor). Write

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

, so

3 x + 1 = A (x - 1) (x + 1) + B (x + 1) + D (x - 1)^{2}

.

- At

x = 1

4 = 2 B

, so

B = 2

- At

x = - 1

- 2 = 4 D

, so

D = - \frac{1}{2}

- Comparing coefficients of

x^{2}

0 = A + D

, so

A = \frac{1}{2}

The integral is

\frac{1}{2} lo g ∣ x - 1∣ - \frac{2}{x - 1} - \frac{1}{2} lo g ∣ x + 1∣ + C

.

An everyday example. Splitting a shared restaurant bill into each person's share turns one complicated total into simple parts, just as partial fractions do.

The substance. A repeated factor needs every power up to its multiplicity — using only

\frac{A}{x - 1}

for

(x - 1)^{2}

can never match the numerator.

How do you split a rational function with an irreducible quadratic factor into partial fractions?

**For a quadratic factor such as

x^{2} + 1

that cannot be factorised over the real numbers, the partial fraction has a linear numerator,

\frac{B x + D}{x ^{2} + 1}

, and it integrates into a logarithm plus an inverse tangent.

Worked example.**

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

Clearing denominators,

2 x + 3 = A (x^{2} + 1) + (B x + D) (x - 1)

.

- At

x = 1

5 = 2 A

, so

A = \frac{5}{2}

- Comparing coefficients of

x^{2}

0 = A + B

, so

B = - \frac{5}{2}

- Comparing constants:

3 = A - D

, so

D = - \frac{1}{2}

Then

\int \frac{2 x + 3}{( x - 1 ) ( x ^{2} + 1 )} d x = \frac{5}{2} lo g ∣ x - 1∣ - \frac{5}{4} lo g (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C

Numerical check. At

x = 2

the fraction equals

\frac{7}{5} = 1.4

, and the partial fractions give

2.5 + \frac{- 5 - 0.5}{5} = 2.5 - 1.1 = 1.4

.

An everyday example. An engineer analysing how a circuit responds to a signal routinely splits such fractions, where the quadratic part corresponds to an oscillation.

The substance. A constant numerator over a quadratic factor is not enough — the numerator must be the general linear form

B x + D

How do you integrate an improper rational function by first reducing it to a proper fraction?

When the degree of the numerator is at least the degree of the denominator, divide first to write the fraction as a polynomial plus a proper fraction, then integrate the polynomial directly and the proper fraction by partial fractions.

Worked example. Find

\int \frac{x ^{3} + x + 1}{x ^{2} - 1} d x

.

- Division:

x^{3} + x + 1 = x (x^{2} - 1) + 2 x + 1

, so the integrand is

x + \frac{2 x + 1}{x ^{2} - 1}

- Partial fractions:

\frac{2 x + 1}{( x - 1 ) ( x + 1 )} = \frac{3/2}{x - 1} + \frac{1/2}{x + 1}

\int \frac{x ^{3} + x + 1}{x ^{2} - 1} d x = \frac{x ^{2}}{2} + \frac{3}{2} lo g ∣ x - 1∣ + \frac{1}{2} lo g ∣ x + 1∣ + C

Numerical check. At

x = 2

the integrand is

\frac{11}{3} \approx 3.667

, and

2 + \frac{3}{2} + \frac{1}{6} \approx 3.667

.

Worked example 2.

\frac{x ^{2}}{x ^{2} + 4} = 1 - \frac{4}{x ^{2} + 4}

, so

\int \frac{x ^{2}}{x ^{2} + 4} d x = x - 2 tan^{- 1} \frac{x}{2} + C

.

An everyday example. Sharing 17 rupees among 5 people as 3 rupees each with 2 left over is the same step — separate the whole part before dealing with the remainder.

The substance. Partial fractions apply only to proper fractions — trying them on an improper fraction leads to equations with no solution.

Exam tip

What earns full marks on integration by parts and partial fractions?

Write the partial fraction form with all unknown constants before solving for them, and in integration by parts state clearly which factor is u.

-

\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x

- Distinct linear factors:

\frac{A}{x - a}

; repeated factors: add

\frac{B}{( x - a ) ^{2}}

- Irreducible quadratic factor: numerator

B x + D

- Improper fractions: divide first

The trap. Using a constant numerator over a quadratic factor. **The numerator must be

B x + D

.**

Did you know

Why do engineers split fractions to understand circuits and vibrations?

Many real systems — electrical circuits, vehicle suspensions and buildings swaying in the wind — are described by fractions of polynomials in a variable that represents frequency.

Splitting such a fraction into partial fractions separates the response into simple parts: linear factors give smoothly fading behaviour, while irreducible quadratic factors give oscillations. Each part can then be studied on its own.

The same algebra from this chapter helps engineers predict whether a bridge will shake or a circuit will hum.

Exam relevance

How are integration by parts and partial fractions tested in JEE Main?

Integrals is a recurring JEE Main chapter, and these two methods, often combined with substitution, appear in both indefinite and definite integral questions.

What gets asked. **Integrals of the form

e^{x} [f (x) + f^{'} (x)]

, products involving logarithms and inverse trigonometric functions, and rational functions that need partial fractions after a substitution.

Question types. Multiple-choice and numerical-value questions, sometimes asking for a constant in the final answer.

The trap that costs marks. Choosing u badly** in integration by parts, which turns a two-line integral into an impossible one.

Key takeaways

What must you be able to do from this lesson?

- Integration by parts:

\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x

, with u chosen to simplify on differentiation
- Linear and repeated factors:

\frac{A}{x - a}

terms, plus

\frac{B}{( x - a ) ^{2}}

for repeated factors
- Quadratic factors:

\frac{B x + D}{x ^{2} + p x + q}

, giving logarithms and inverse tangents
- Improper fractions: divide first, then use partial fractions

Can you find

\int x lo g x d x

using integration by parts?

What do you do when an integrand is a product or a fraction?

How do you integrate a product of functions using integration by parts?

How do you split a proper rational function with linear and repeated linear factors into partial fractions?

How do you split a rational function with an irreducible quadratic factor into partial fractions?

How do you integrate an improper rational function by first reducing it to a proper fraction?

What earns full marks on integration by parts and partial fractions?

Why do engineers split fractions to understand circuits and vibrations?

How are integration by parts and partial fractions tested in JEE Main?

What must you be able to do from this lesson?

Related study notes

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