NCERT Class 12 Mathematics — Integrals, Exercise 7.4. All 25 questions solved.

Exercise 7.4 works from a fixed list of standard forms. Almost nothing here is invented; the skill is recognising which form an integral belongs to and manipulating it until it fits.

$$\int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + \mathrm{C}, \qquad \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + \mathrm{C}$$

$$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + \mathrm{C}, \qquad \int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}\frac{x}{a} + \mathrm{C}$$

$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2+a^2}\right| + \mathrm{C}, \qquad \int \frac{dx}{\sqrt{x^2 – a^2}} = \log\left|x + \sqrt{x^2-a^2}\right| + \mathrm{C}$$

Two manipulations get almost every question into one of these shapes.

Key insight. Everything in this exercise is one of two moves.

Complete the square when the denominator is a quadratic: $x^2 + 2x + 2 = (x+1)^2 + 1$, and now it is a standard form in the variable $x+1$. Questions 10 to 15 are all this.

Split a linear numerator when one is present. Write it as $\lambda\big(\text{derivative of the quadratic}\big) + \mu$. The $\lambda$ part integrates by substitution, the $\mu$ part by a standard form. Questions 16 to 23 are all this, and there is no way through them without it.

Every answer has been checked by differentiating it back to the integrand.

Questions 1 to 23 — integrate the functions.

Question 1

$\dfrac{3x^2}{x^6 + 1}$

Solution. Write $x^6 = (x^3)^2$, and note $3x^2$ is the derivative of $x^3$. Putting $t = x^3$:

$$\int \frac{dt}{t^2 + 1} = \tan^{-1}t$$

Question 2

$\dfrac{1}{\sqrt{1 + 4x^2}}$

Solution. Put $t = 2x$, so $dt = 2\,dx$, and the form becomes $\tfrac{1}{\sqrt{t^2 + 1}}$:

$$\frac{1}{2}\int \frac{dt}{\sqrt{t^2+1}} = \frac{1}{2}\log\left|t + \sqrt{t^2+1}\right|$$

Question 3

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

Solution. Put $t = 2 – x$, so $dt = -dx$ — the minus sign flips the logarithm’s argument:

$$-\int \frac{dt}{\sqrt{t^2+1}} = -\log\left|t + \sqrt{t^2+1}\right| = \log\left|\frac{1}{t + \sqrt{t^2+1}}\right|$$

Question 4

$\dfrac{1}{\sqrt{9 – 25x^2}}$

Solution. Factor out the 25 to reach the $\sqrt{a^2 – t^2}$ form:

$$\sqrt{9 – 25x^2} = 5\sqrt{\tfrac{9}{25} – x^2}$$

$$\frac{1}{5}\int \frac{dx}{\sqrt{\left(\frac{3}{5}\right)^2 – x^2}} = \frac{1}{5}\sin^{-1}\frac{x}{3/5}$$

Question 5

$\dfrac{3x}{1 + 2x^4}$

Solution. $x^4 = (x^2)^2$ and $x\,dx$ is a multiple of $d(x^2)$. Put $t = x^2$, $dt = 2x\,dx$:

$$\frac{3}{2}\int \frac{dt}{1 + 2t^2} = \frac{3}{4}\int \frac{dt}{\frac{1}{2} + t^2} = \frac{3}{4}\cdot\sqrt{2}\tan^{-1}\left(\sqrt{2}\,t\right)$$

Question 6

$\dfrac{x^2}{1 – x^6}$

Solution. With $t = x^3$, $dt = 3x^2 dx$, this is the $\tfrac{1}{a^2-t^2}$ form with $a = 1$:

$$\frac{1}{3}\int \frac{dt}{1 – t^2} = \frac{1}{3}\cdot\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|$$

Question 7

$\dfrac{x – 1}{\sqrt{x^2 – 1}}$

Solution. Split the numerator — the first piece is a substitution, the second a standard form:

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

For the first, $t = x^2 – 1$ gives $\sqrt{x^2-1}$; the second is the standard logarithm.

Question 8

$\dfrac{x^2}{\sqrt{x^6 + a^6}}$

Solution. Again $t = x^3$, $dt = 3x^2 dx$, giving the $\tfrac{1}{\sqrt{t^2 + (a^3)^2}}$ form:

$$\frac{1}{3}\int \frac{dt}{\sqrt{t^2 + \left(a^3\right)^2}} = \frac{1}{3}\log\left|t + \sqrt{t^2 + a^6}\right|$$

Question 9

$\dfrac{\sec^2 x}{\sqrt{\tan^2 x + 4}}$

Solution. $\sec^2 x$ is the derivative of $\tan x$, so $t = \tan x$ turns this into a standard form with $a = 2$:

$$\int \frac{dt}{\sqrt{t^2 + 4}} = \log\left|t + \sqrt{t^2+4}\right|$$

Question 10

$\dfrac{1}{\sqrt{x^2 + 2x + 2}}$

Solution. Complete the square. Half of the $x$ coefficient is 1, so:

$$x^2 + 2x + 2 = (x+1)^2 + 1$$

With $t = x + 1$ this is the standard $\tfrac{1}{\sqrt{t^2+1}}$ form.

Question 11

$\dfrac{1}{9x^2 + 6x + 5}$

Solution. Complete the square, treating $9x^2 + 6x$ as $(3x)^2 + 2(3x)$:

$$9x^2 + 6x + 5 = (3x + 1)^2 + 4$$

With $t = 3x + 1$, $dt = 3\,dx$, and $a = 2$:

$$\frac{1}{3}\int \frac{dt}{t^2 + 4} = \frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2}$$

Question 12

$\dfrac{1}{\sqrt{7 – 6x – x^2}}$

Solution. The $x^2$ is negative, so factor out the minus sign before completing the square:

$$7 – 6x – x^2 = 7 – \left(x^2 + 6x\right) = 7 + 9 – (x+3)^2 = 16 – (x+3)^2$$

This is the $\sqrt{a^2 – t^2}$ form with $a = 4$, $t = x+3$.

Question 13

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

Solution. Multiply out first, then complete the square:

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

With $t = x – \tfrac32$ and $a = \tfrac12$, this is the $\sqrt{t^2 – a^2}$ form.

Question 14

$\dfrac{1}{\sqrt{8 + 3x – x^2}}$

Solution.

$$8 + 3x – x^2 = 8 – \left(x^2 – 3x\right) = 8 + \frac{9}{4} – \left(x – \frac{3}{2}\right)^2 = \frac{41}{4} – \left(x – \frac{3}{2}\right)^2$$

So $a = \tfrac{\sqrt{41}}{2}$ and $t = x – \tfrac32$:

$$\sin^{-1}\frac{x – \frac{3}{2}}{\frac{\sqrt{41}}{2}} = \sin^{-1}\frac{2x – 3}{\sqrt{41}}$$

Question 15

$\dfrac{1}{\sqrt{(x-a)(x-b)}}$

Solution. The general version of question 13:

$$(x-a)(x-b) = x^2 – (a+b)x + ab = \left(x – \frac{a+b}{2}\right)^2 – \left(\frac{a-b}{2}\right)^2$$

Question 16

$\dfrac{4x + 1}{\sqrt{2x^2 + x – 3}}$

Solution. Here the numerator-splitting method begins. Note $\tfrac{d}{dx}\left(2x^2 + x – 3\right) = 4x + 1$ — the numerator is exactly the derivative, so no splitting is needed at all.

With $t = 2x^2 + x – 3$:

$$\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}$$

Question 17

$\dfrac{x + 2}{\sqrt{x^2 – 1}}$

Solution. $\tfrac{d}{dx}(x^2-1) = 2x$, so write $x + 2 = \tfrac12(2x) + 2$ and split:

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

The first gives $\sqrt{x^2-1}$; the second is the standard logarithm.

Question 18

$\dfrac{5x – 2}{1 + 2x + 3x^2}$

Solution. $\tfrac{d}{dx}\left(3x^2 + 2x + 1\right) = 6x + 2$. Write

$$5x – 2 = \frac{5}{6}(6x + 2) – \frac{11}{3}$$

(checking: $\tfrac56 \cdot 2 = \tfrac53$, and $\tfrac53 – \tfrac{11}{3} = -2$ ✓). The first part gives a logarithm by substitution. For the second, complete the square:

$$3x^2 + 2x + 1 = 3\left[\left(x + \frac{1}{3}\right)^2 + \frac{2}{9}\right]$$

giving an arctangent with $t = 3x+1$ after simplification.

Question 19

$\dfrac{6x + 7}{\sqrt{(x-5)(x-4)}}$

Solution. Expand: $(x-5)(x-4) = x^2 – 9x + 20$, whose derivative is $2x – 9$. Split:

$$6x + 7 = 3(2x – 9) + 34$$

The first part gives $6\sqrt{x^2-9x+20}$; the second needs the square completed, $x^2 – 9x + 20 = \left(x – \tfrac92\right)^2 – \tfrac14$.

Question 20

$\dfrac{x + 2}{\sqrt{4x – x^2}}$

Solution. $\tfrac{d}{dx}\left(4x – x^2\right) = 4 – 2x$, so write

$$x + 2 = -\frac{1}{2}(4 – 2x) + 4$$

The first part integrates to $-\sqrt{4x-x^2}$. For the second, complete the square: $4x – x^2 = 4 – (x-2)^2$, giving an arcsine with $a = 2$.

Question 21

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

Solution. The derivative of the quadratic is $2x + 2$, so

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

First part: $\sqrt{x^2+2x+3}$. Second: complete the square, $x^2 + 2x + 3 = (x+1)^2 + 2$.

Question 22

$\dfrac{x + 3}{x^2 – 2x – 5}$

Solution. Derivative of the denominator is $2x – 2$, so

$$x + 3 = \frac{1}{2}(2x – 2) + 4$$

The first part gives $\tfrac12\log\left|x^2-2x-5\right|$. For the second, $x^2 – 2x – 5 = (x-1)^2 – 6$, which is the $\tfrac{1}{t^2-a^2}$ form with $a = \sqrt{6}$:

$$4 \cdot \frac{1}{2\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|$$

Question 23

$\dfrac{5x + 3}{\sqrt{x^2 + 4x + 10}}$

Solution. Derivative of the quadratic is $2x + 4$, so

$$5x + 3 = \frac{5}{2}(2x + 4) – 7$$

(checking: $\tfrac52 \times 4 = 10$, and $10 – 7 = 3$ ✓). The first part gives $5\sqrt{x^2+4x+10}$. For the second, $x^2 + 4x + 10 = (x+2)^2 + 6$.

Questions 24 and 25 — choose the correct answer.

Question 24

$\displaystyle\int \frac{dx}{x^2 + 2x + 2}$ equals

Solution. Complete the square:

$$x^2 + 2x + 2 = (x+1)^2 + 1$$

With $t = x + 1$ and $a = 1$, the standard form gives $\tan^{-1}t$ directly — no extra factor of $x$ appears, which rules out (A) and (C).

Question 25

$\displaystyle\int \frac{dx}{\sqrt{9x – 4x^2}}$ equals

Solution. Factor out the 4 before completing the square, since the coefficient of $x^2$ is not 1:

$$9x – 4x^2 = -4\left(x^2 – \frac{9x}{4}\right) = -4\left[\left(x – \frac{9}{8}\right)^2 – \frac{81}{64}\right]$$

$$= 4\left[\left(\frac{9}{8}\right)^2 – \left(x – \frac{9}{8}\right)^2\right]$$

The $\sqrt{4} = 2$ comes outside as $\tfrac12$, and the standard arcsine form with $a = \tfrac98$ gives

$$\frac{1}{2}\sin^{-1}\frac{x – \frac{9}{8}}{\frac{9}{8}} = \frac{1}{2}\sin^{-1}\left(\frac{8x – 9}{9}\right)$$

Option (D) has the $8$ and $9$ transposed inside the arcsine — the arithmetic of $\tfrac{x – 9/8}{9/8}$ is exactly what separates the two.

Common mistakes

• Forgetting to factor out the leading coefficient before completing the square. In questions 4, 11 and 25 the $x^2$ term has a coefficient, and it must come out first. Question 25’s distractors are built from getting this wrong.
• Mishandling the sign when $x^2$ is negative. In questions 12, 14 and 20 the quadratic opens downward, so pull out the minus sign first and expect an arcsine, not a logarithm.
• Splitting the numerator with the wrong constants. Always check: if $px + q = \lambda\big(\text{derivative}\big) + \mu$, substitute back and confirm both coefficients. Question 23’s $\tfrac52$ and $-7$ are easy to get slightly wrong.
• Confusing the two logarithm forms. $\int\tfrac{dx}{x^2-a^2}$ gives $\tfrac{1}{2a}\log\left|\tfrac{x-a}{x+a}\right|$, while $\int\tfrac{dx}{\sqrt{x^2-a^2}}$ gives $\log\left|x+\sqrt{x^2-a^2}\right|$. The square root changes the answer completely.
• Missing when the numerator is already the derivative. Question 16 needs no splitting at all — $4x+1$ is exactly $\tfrac{d}{dx}\left(2x^2+x-3\right)$. Check for this before doing any algebra.

Practise next

• Exercise 7.5 — integration by partial fractions, for rational functions that these standard forms cannot reach.
• Exercise 7.3 — worth revisiting if completing the square still feels slow.