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CBSE Board Examination 2026 · Section C · 3 marks

Q27: Find ∫x+2x−2 dx\int\sqrt{\frac{x + 2}{x - 2}}\,dx⁠, or ∫x2(x2+9)(x2+16) dx\int\frac{x^2}{(x^2 + 9)(x^2 + 16)}\,dx

Board 2026Ch 7 · Integrals
The question

(a) Find ∫x+2x−2 dx\displaystyle\int\sqrt{\frac{x + 2}{x - 2}}\,dx

OR

(b) Find : ∫x2(x2+9)(x2+16) dx\displaystyle\int\frac{x^2}{(x^2 + 9)(x^2 + 16)}\,dx

Exam answer

(a)

Let I=∫x+2x−2 dxI = \displaystyle\int\sqrt{\frac{x + 2}{x - 2}}\,dx⁠, x>2x > 2⁠.

Multiplying the numerator and the denominator under the root by x+2x + 2⁠,

I=∫(x+2)2x2−4 dx=∫x+2x2−4 dx=12∫2xx2−4 dx+2∫1x2−4 dx=I1+I2\begin{aligned} I &= \int\sqrt{\frac{(x + 2)^2}{x^2 - 4}}\,dx \\ &= \int\frac{x + 2}{\sqrt{x^2 - 4}}\,dx \\ &= \frac12\int\frac{2x}{\sqrt{x^2 - 4}}\,dx \\ &\quad + 2\int\frac{1}{\sqrt{x^2 - 4}}\,dx \\ &= I_1 + I_2 \end{aligned}
Why?

A square root of a fraction has no standard integral. Multiplying top and bottom by x+2x + 2 makes the top a perfect square, which comes out of the root. On the bottom,

(x+2)(x−2)=x2−4(x + 2)(x - 2) = x^2 - 4 x+2x−2=(x+2)(x+2)(x−2)(x+2)=(x+2)2x2−4\begin{aligned} \frac{x + 2}{x - 2} &= \frac{(x + 2)(x + 2)}{(x - 2)(x + 2)} \\ &= \frac{(x + 2)^2}{x^2 - 4} \end{aligned}

For x>2x > 2⁠, x+2>0x + 2 > 0⁠, so (x+2)2=x+2\sqrt{(x + 2)^2} = x + 2⁠.

(x+2)2x2−4=(x+2)2x2−4=x+2x2−4\begin{aligned} \sqrt{\frac{(x + 2)^2}{x^2 - 4}} &= \frac{\sqrt{(x + 2)^2}}{\sqrt{x^2 - 4}} \\ &= \frac{x + 2}{\sqrt{x^2 - 4}} \end{aligned}

Then split the top: x+2=12(2x)+2x + 2 = \frac12(2x) + 2⁠. The 2x2x is the derivative of x2−4x^2 - 4⁠, which suits a substitution. The 2 leaves a standard form.

x+2x2−4=12(2x)+2x2−4=12⋅2xx2−4+2⋅1x2−4\begin{aligned} \frac{x + 2}{\sqrt{x^2 - 4}} &= \frac{\frac12(2x) + 2}{\sqrt{x^2 - 4}} \\ &= \frac12\cdot\frac{2x}{\sqrt{x^2 - 4}} \\ &\quad + 2\cdot\frac{1}{\sqrt{x^2 - 4}} \end{aligned}

In I1I_1⁠, put x2−4=tx^2 - 4 = t⁠. Then 2x dx=dt2x\,dx = dt⁠.

I1=12∫dtt=12⋅2t=x2−4\begin{aligned} I_1 &= \frac12\int\frac{dt}{\sqrt t} = \frac12\cdot 2\sqrt t \\ &= \sqrt{x^2 - 4} \end{aligned}
Why?

The substitution replaces 2x dx2x\,dx by dtdt and x2−4x^2 - 4 by tt⁠.

By the power rule with n=−12n = -\frac12⁠,

∫t−1/2 dt=t1/21/2=2t\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} = 2\sqrt t

Then put t=x2−4t = x^2 - 4 back. One constant CC is added at the end.

Using ∫dxx2−a2\displaystyle\int\frac{dx}{\sqrt{x^2 - a^2}} =log⁡∣x+x2−a2∣+C= \log\left|x + \sqrt{x^2 - a^2}\right| + C⁠,

I2=2log⁡∣x+x2−4∣I_2 = 2\log\left|x + \sqrt{x^2 - 4}\right|
Why?

x2−4=x2−22x^2 - 4 = x^2 - 2^2⁠, so this is the standard form with a=2a = 2⁠. The factor 2 in front of I2I_2 stays.

Watch out: 1x2−a2\frac{1}{\sqrt{x^2 - a^2}} gives a log, while 1a2−x2\frac{1}{\sqrt{a^2 - x^2}} gives sin⁡−1xa\sin^{-1}\frac{x}{a}⁠.

∴ I=x2−4\therefore\ I = \sqrt{x^2 - 4} +2log⁡∣x+x2−4∣+C+ 2\log\left|x + \sqrt{x^2 - 4}\right| + C⁠.

(b)

Put x2=tx^2 = t (for the partial fractions only):

t(t+9)(t+16)=At+9+Bt+16\frac{t}{(t + 9)(t + 16)} = \frac{A}{t + 9} + \frac{B}{t + 16}
Why?

xx appears only as x2x^2⁠, so writing tt for x2x^2 makes the factors linear and the partial fractions easy. This is not a change of variable in the integral: dxdx stays as it is.

The degree of the top (1 in tt⁠) is less than that of the bottom (2), and the bottom has two different linear factors, so each factor gets a constant over it.

⇒t=A(t+16)+B(t+9)\Rightarrow t = A(t + 16) + B(t + 9)
Why?

Multiply both sides by (t+9)(t+16)(t + 9)(t + 16)⁠. This must hold for every value of tt⁠, so we may choose convenient values.

Putting t=−9t = -9⁠: −9=7A-9 = 7A⁠, so A=−97A = -\frac97⁠.

Putting t=−16t = -16⁠: −16=−7B-16 = -7B⁠, so B=167B = \frac{16}{7}⁠.

Why?

t=−9t = -9 makes the BB term zero, leaving A(−9+16)=7AA(-9 + 16) = 7A⁠.

−9=A(−9+16)+B(−9+9)−9=A(7)+B(0)−9=7AA=−97\begin{aligned} -9 &= A(-9 + 16) + B(-9 + 9) \\ -9 &= A(7) + B(0) \\ -9 &= 7A \\ A &= -\frac{9}{7} \end{aligned}

t=−16t = -16 makes the AA term zero, leaving B(−16+9)=−7BB(-16 + 9) = -7B⁠.

−16=A(−16+16)+B(−16+9)−16=A(0)+B(−7)−16=−7BB=−16−7=167\begin{aligned} -16 &= A(-16 + 16) + B(-16 + 9) \\ -16 &= A(0) + B(-7) \\ -16 &= -7B \\ B &= \frac{-16}{-7} = \frac{16}{7} \end{aligned}
∴ ∫x2(x2+9)(x2+16) dx=−97∫dxx2+32+167∫dxx2+42\begin{aligned} \therefore\ &\int\frac{x^2}{(x^2 + 9)(x^2 + 16)}\,dx \\ &= -\frac97\int\frac{dx}{x^2 + 3^2} \\ &\quad + \frac{16}{7}\int\frac{dx}{x^2 + 4^2} \end{aligned}
Why?

Put t=x2t = x^2 back, and write 9=329 = 3^2 and 16=4216 = 4^2 to show the standard form x2+a2x^2 + a^2⁠.

x2(x2+9)(x2+16)=−97x2+9+167x2+16=−97⋅1x2+32+167⋅1x2+42\begin{aligned} &\frac{x^2}{(x^2 + 9)(x^2 + 16)} \\ &= \frac{-\frac97}{x^2 + 9} + \frac{\frac{16}{7}}{x^2 + 16} \\ &= -\frac97\cdot\frac{1}{x^2 + 3^2} \\ &\quad + \frac{16}{7}\cdot\frac{1}{x^2 + 4^2} \end{aligned}

Using ∫dxx2+a2=1atan⁡−1xa+C\displaystyle\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + C⁠,

=−97⋅13tan⁡−1x3+167⋅14tan⁡−1x4+C\begin{aligned} &= -\frac97\cdot\frac13\tan^{-1}\frac{x}{3} \\ &\quad + \frac{16}{7}\cdot\frac14\tan^{-1}\frac{x}{4} + C \end{aligned}
Why?

The first integral has a=3a = 3 and the second has a=4a = 4⁠.

Then 97⋅13=37\frac97\cdot\frac13 = \frac37 and 167⋅14=47\frac{16}{7}\cdot\frac14 = \frac47⁠.

97⋅13=921=9÷321÷3=37167⋅14=1628=16÷428÷4=47\begin{aligned} \frac97\cdot\frac13 &= \frac{9}{21} = \frac{9 \div 3}{21 \div 3} = \frac37 \\ \frac{16}{7}\cdot\frac14 &= \frac{16}{28} = \frac{16 \div 4}{28 \div 4} = \frac47 \end{aligned}

∴\therefore The integral =−37tan⁡−1x3= -\dfrac37\tan^{-1}\dfrac{x}{3} +47tan⁡−1x4+C+ \dfrac47\tan^{-1}\dfrac{x}{4} + C⁠.

In (a), the working takes x>2x > 2⁠, as the marking scheme does. For x<−2x < -2 the integrand equals −x+2x2−4-\frac{x + 2}{\sqrt{x^2 - 4}}⁠, so the answer there is the negative of the same expression, plus CC⁠.

Where this comes from

Chapter 7, Integrals: Integrals by rationalising; partial fractions.

Question type: (a) A linear top over a quadratic; (b) Partial fractions

How to spot it: click to see

(a) Type 5. A linear top over a quadratic: split the top

Spot it: px+qpx + q on top, and a quadratic, or the square root of one, at the bottom.

Here: Multiplying top and bottom under the root by x+2x + 2 gives x+2x2−4\frac{x + 2}{\sqrt{x^2 - 4}}⁠. Write x+2=12(2x)+2x + 2 = \frac12(2x) + 2⁠: the 2x2x part gives x2−4\sqrt{x^2 - 4}⁠, the 2 part a log card.

(b) Type 7. Partial fractions: break a fraction into simple pieces

Spot it: a fraction of two polynomials whose bottom splits into factors.

Here: The bottom (x2+9)(x2+16)(x^2 + 9)(x^2 + 16) has two different factors, and xx appears only as x2x^2⁠. Write tt for x2x^2 to find A=−97A = -\frac97 and B=167B = \frac{16}{7}⁠; each piece is then a tan⁡−1\tan^{-1} card.

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