NCERT Class 12 Mathematics — Integrals, Exercise 7.5. All 23 questions solved.

Exercise 7.5 is integration by partial fractions: splitting a rational function into pieces simple enough to integrate. The form of the split is decided entirely by the factors of the denominator.

$$\frac{px+q}{(x-a)(x-b)} = \frac{\mathrm A}{x-a} + \frac{\mathrm B}{x-b}, \qquad \frac{px+q}{(x-a)^2} = \frac{\mathrm A}{x-a} + \frac{\mathrm B}{(x-a)^2}$$

$$\frac{px^2+qx+r}{(x-a)\left(x^2+bx+c\right)} = \frac{\mathrm A}{x-a} + \frac{\mathrm Bx + \mathrm C}{x^2+bx+c}$$

with the last form used only when the quadratic cannot be factorised.

Key insight. Two things have to be true before you decompose, and skipping either is what goes wrong.

The fraction must be proper — the numerator’s degree strictly less than the denominator’s. If it is not, divide first. Question 12 is improper and gives an $\tfrac{x^2}{2}$ term that no decomposition would ever produce.

Every repeated factor needs every power. $(x-1)^2$ contributes both $\tfrac{\mathrm A}{x-1}$ and $\tfrac{\mathrm B}{(x-1)^2}$, not just the square. Question 8 turns on this.

The fastest way to find the constants is to substitute the roots of the denominator, which kills all but one term at a time.

Questions 1 to 21 — integrate the rational functions.

Question 1

$\dfrac{x}{(x+1)(x+2)}$

Solution. Write $\dfrac{x}{(x+1)(x+2)} = \dfrac{\mathrm A}{x+1} + \dfrac{\mathrm B}{x+2}$, so $x = \mathrm A(x+2) + \mathrm B(x+1)$.

Putting $x = -1$: $-1 = \mathrm A$. Putting $x = -2$: $-2 = -\mathrm B$, so $\mathrm B = 2$.

$$\int \left(\frac{-1}{x+1} + \frac{2}{x+2}\right)dx = -\log|x+1| + 2\log|x+2|$$

Question 2

$\dfrac{1}{x^2 – 9}$

Solution. This is the standard form $\tfrac{1}{x^2-a^2}$ with $a = 3$, which is itself a partial-fraction result:

$$\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|$$

Question 3

$\dfrac{3x-1}{(x-1)(x-2)(x-3)}$

Solution. Three distinct linear factors, so three terms. Substituting each root in $3x – 1 = \mathrm A(x-2)(x-3) + \mathrm B(x-1)(x-3) + \mathrm C(x-1)(x-2)$:

$$x=1:\ 2 = 2\mathrm A \Rightarrow \mathrm A = 1, \qquad x=2:\ 5 = -\mathrm B \Rightarrow \mathrm B = -5, \qquad x=3:\ 8 = 2\mathrm C \Rightarrow \mathrm C = 4$$

Question 4

$\dfrac{x}{(x-1)(x-2)(x-3)}$

Solution. Same three factors, different numerator:

$$x=1:\ 1 = 2\mathrm A \Rightarrow \mathrm A = \tfrac12, \qquad x=2:\ 2 = -\mathrm B \Rightarrow \mathrm B = -2, \qquad x=3:\ 3 = 2\mathrm C \Rightarrow \mathrm C = \tfrac32$$

Question 5

$\dfrac{2x}{x^2 + 3x + 2}$

Solution. Factor the denominator first: $x^2 + 3x + 2 = (x+1)(x+2)$.

$$x=-1:\ -2 = \mathrm A \Rightarrow \mathrm A = -2, \qquad x=-2:\ -4 = -\mathrm B \Rightarrow \mathrm B = 4$$

Question 6

$\dfrac{1 – x^2}{x(1 – 2x)}$

Solution. Degree 2 over degree 2 — improper, so divide first:

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

Decomposing the remainder gives $\tfrac{1}{x} – \tfrac{3}{2(1-2x)}$, and integrating term by term:

Question 7

$\dfrac{x}{(x^2+1)(x-1)}$

Solution. The quadratic $x^2+1$ does not factorise, so it takes a linear numerator:

$$\frac{x}{(x^2+1)(x-1)} = \frac{\mathrm A}{x-1} + \frac{\mathrm Bx + \mathrm C}{x^2+1}$$

Putting $x = 1$ gives $1 = 2\mathrm A$, so $\mathrm A = \tfrac12$; comparing coefficients then gives $\mathrm B = -\tfrac12$, $\mathrm C = \tfrac12$.

The $\tfrac{Bx}{x^2+1}$ piece integrates to a logarithm and the constant piece to an arctangent:

Question 8

$\dfrac{x}{(x-1)^2(x+2)}$

Solution. The repeated factor needs both powers:

$$\frac{\mathrm A}{x-1} + \frac{\mathrm B}{(x-1)^2} + \frac{\mathrm C}{x+2}$$

Putting $x = 1$ gives $1 = 3\mathrm B$, so $\mathrm B = \tfrac13$; $x = -2$ gives $-2 = 9\mathrm C$, so $\mathrm C = -\tfrac29$; comparing $x^2$ coefficients gives $\mathrm A = \tfrac29$.

Question 9

$\dfrac{3x+5}{x^3 – x^2 – x + 1}$

Solution. Factor the cubic by grouping — this is the real work of the question:

$$x^3 – x^2 – x + 1 = x^2(x-1) – (x-1) = (x-1)\left(x^2-1\right) = (x-1)^2(x+1)$$

So again there is a repeated factor. Solving gives $\mathrm A = -\tfrac12$, $\mathrm B = 4$, $\mathrm C = \tfrac12$ in $\tfrac{\mathrm A}{x-1} + \tfrac{\mathrm B}{(x-1)^2} + \tfrac{\mathrm C}{x+1}$.

Question 10

$\dfrac{2x-3}{(x^2-1)(2x+3)}$

Solution. $x^2 – 1 = (x-1)(x+1)$, giving three distinct linear factors. Substituting the roots $x = 1$, $x = -1$ and $x = -\tfrac32$:

$$\mathrm A = -\frac{1}{10}, \qquad \mathrm B = \frac{5}{2}, \qquad \mathrm C = -\frac{24}{5}$$

for $\tfrac{\mathrm A}{x-1} + \tfrac{\mathrm B}{x+1} + \tfrac{\mathrm C}{2x+3}$. The last term integrates with an extra $\tfrac12$ from the $2x$.

Question 11

$\dfrac{5x}{(x+1)(x^2-4)}$

Solution. $x^2 – 4 = (x-2)(x+2)$, so the roots are $-1$, $2$ and $-2$:

$$\mathrm A = \frac{5}{3}, \qquad \mathrm B = \frac{5}{6}, \qquad \mathrm C = -\frac{5}{2}$$

Question 12

$\dfrac{x^3 + x + 1}{x^2 – 1}$

Solution. Degree 3 over degree 2 — improper, so divide:

$$\frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1}$$

Now decompose the proper remainder over $(x-1)(x+1)$: $\mathrm A = \tfrac32$, $\mathrm B = \tfrac12$.

$$\int \left(x + \frac{3/2}{x-1} + \frac{1/2}{x+1}\right)dx$$

Question 13

$\dfrac{2}{(1-x)(1+x^2)}$

Solution. One linear factor and one irreducible quadratic:

$$\frac{\mathrm A}{1-x} + \frac{\mathrm Bx + \mathrm C}{1+x^2}$$

Putting $x = 1$ gives $\mathrm A = 1$; comparison gives $\mathrm B = \mathrm C = 1$.

Question 14

$\dfrac{3x-1}{(x+2)^2}$

Solution. A repeated factor with nothing else. Write the numerator in terms of $x+2$: $3x – 1 = 3(x+2) – 7$.

$$\int \left(\frac{3}{x+2} – \frac{7}{(x+2)^2}\right)dx = 3\log|x+2| + \frac{7}{x+2}$$

Question 15

$\dfrac{1}{x^4 – 1}$

Solution. Factor as far as possible over the reals:

$$x^4 – 1 = \left(x^2-1\right)\left(x^2+1\right) = (x-1)(x+1)\left(x^2+1\right)$$

The decomposition gives $\tfrac{1}{4(x-1)} – \tfrac{1}{4(x+1)} – \tfrac{1}{2(x^2+1)}$.

Question 16

$\dfrac{1}{x(x^n+1)}$

Solution. Follow the hint: multiply top and bottom by $x^{n-1}$ so that a substitution becomes available.

$$\frac{x^{n-1}}{x^n\left(x^n+1\right)}$$

Now $t = x^n$ gives $dt = nx^{n-1}dx$:

$$\frac{1}{n}\int \frac{dt}{t(t+1)} = \frac{1}{n}\int \left(\frac{1}{t} – \frac{1}{t+1}\right)dt = \frac{1}{n}\log\left|\frac{t}{t+1}\right|$$

Question 17

$\dfrac{\cos x}{(1-\sin x)(2-\sin x)}$

Solution. Following the hint, $t = \sin x$ gives $dt = \cos x\,dx$ and turns this into a rational function:

$$\int \frac{dt}{(1-t)(2-t)}$$

Decomposing: $\tfrac{1}{1-t} – \tfrac{1}{2-t}$, which integrates to $-\log|1-t| + \log|2-t|$.

Question 18

$\dfrac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}$

Solution. Degree 4 over degree 4, so improper — divide first. Then treat $x^2$ as the variable when decomposing:

$$= 1 – \frac{4x^2 + 10}{\left(x^2+3\right)\left(x^2+4\right)} = 1 + \frac{2}{x^2+3} – \frac{6}{x^2+4}$$

Each remaining term is a standard arctangent.

Question 19

$\dfrac{2x}{\left(x^2+1\right)\left(x^2+3\right)}$

Solution. Only even powers of $x$ appear apart from the $2x$, so substitute $t = x^2$, $dt = 2x\,dx$:

$$\int \frac{dt}{(t+1)(t+3)} = \frac{1}{2}\int \left(\frac{1}{t+1} – \frac{1}{t+3}\right)dt$$

Question 20

$\dfrac{1}{x\left(x^4-1\right)}$

Solution. Multiply top and bottom by $x^3$ and substitute $t = x^4$, $dt = 4x^3 dx$:

$$\frac{1}{4}\int \frac{dt}{t(t-1)} = \frac{1}{4}\int \left(\frac{1}{t-1} – \frac{1}{t}\right)dt = \frac{1}{4}\log\left|\frac{t-1}{t}\right|$$

Question 21

$\dfrac{1}{e^x – 1}$

Solution. Following the hint, $t = e^x$ gives $dt = e^x dx = t\,dx$, so $dx = \tfrac{dt}{t}$:

$$\int \frac{dt}{t(t-1)} = \int \left(\frac{1}{t-1} – \frac{1}{t}\right)dt = \log\left|\frac{t-1}{t}\right|$$

Questions 22 and 23 — choose the correct answer.

Question 22

$\displaystyle\int \frac{x\,dx}{(x-1)(x-2)}$ equals

Solution. Decompose: $x = \mathrm A(x-2) + \mathrm B(x-1)$.

$$x = 1:\ 1 = -\mathrm A \Rightarrow \mathrm A = -1, \qquad x = 2:\ 2 = \mathrm B$$

$$\int \left(\frac{-1}{x-1} + \frac{2}{x-2}\right)dx = -\log|x-1| + 2\log|x-2|$$

Combining the logarithms puts the squared factor on top:

$$= \log\left|\frac{(x-2)^2}{x-1}\right|$$

Option (A) has the two factors the wrong way round — which is what you get by swapping $\mathrm A$ and $\mathrm B$.

Question 23

$\displaystyle\int \frac{dx}{x\left(x^2+1\right)}$ equals

Solution. The quadratic is irreducible, so:

$$\frac{1}{x\left(x^2+1\right)} = \frac{1}{x} – \frac{x}{x^2+1}$$

which can be checked by recombining. Integrating, the second term is $\tfrac{f’}{f}$ up to a factor of $\tfrac12$:

$$\log|x| – \frac{1}{2}\log\left(x^2+1\right)$$

The minus sign is the whole question — options (B) and (C) differ from the answer only in signs.

Common mistakes

• Decomposing an improper fraction. Questions 6, 12 and 18 must be divided first. A decomposition of an improper fraction gives inconsistent equations and no solution.
• Giving a repeated factor only its highest power. $(x-1)^2$ needs both $\tfrac{\mathrm A}{x-1}$ and $\tfrac{\mathrm B}{(x-1)^2}$. Questions 8, 9 and 14 each have one.
• Putting a constant over an irreducible quadratic. It needs a linear numerator $\mathrm Bx + \mathrm C$, which then splits into a logarithm and an arctangent. Questions 7, 13 and 15.
• Forgetting to factor the denominator first. Questions 5, 9, 10, 11 and 15 all look harder than they are until the denominator is factorised — question 9’s cubic factors by grouping into $(x-1)^2(x+1)$.
• Missing the substitution route. Questions 16, 19, 20 and 21 become rational only after a substitution; decomposing them directly in $x$ is far harder.
• Losing the extra constant from a non-unit coefficient. $\int\tfrac{dx}{2x+3} = \tfrac12\log|2x+3|$, not $\log|2x+3|$. Question 10.

Practise next

• Exercise 7.6 — integration by parts, the last of the four main techniques.
• Exercise 7.4 — worth revisiting, since several answers here end in the same standard forms.