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Q.(a) Find: ∫x+2x−2 dx\displaystyle\int \sqrt{\dfrac{x+2}{x-2}}\,dx.

(OR)
(b) Find: ∫x2(x2+9)(x2+16) dx\displaystyle\int \dfrac{x^2}{(x^2 + 9)(x^2 + 16)}\,dx.
CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★
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  1. ∫x+2x−2 dx=x2−4+2ln⁡∣x+x2−4∣+C\displaystyle\int\sqrt{\tfrac{x+2}{x-2}}\,dx=\sqrt{x^2-4}+2\ln\left|x+\sqrt{x^2-4}\right|+C.
  2. ∫x2(x2+9)(x2+16) dx=47tan⁡−1x4−37tan⁡−1x3+C\displaystyle\int\frac{x^2}{(x^2+9)(x^2+16)}\,dx=\frac47\tan^{-1}\frac x4-\frac37\tan^{-1}\frac x3+C.

Part (a)

The square root of a ratio is awkward, so rationalise by multiplying inside by (x+2)(x+2):

x+2x−2=x+2x−2⋅x+2x+2=x+2(x+2)(x−2)=x+2x2−4.\sqrt{\frac{x+2}{x-2}}=\frac{\sqrt{x+2}}{\sqrt{x-2}}\cdot\frac{\sqrt{x+2}}{\sqrt{x+2}}=\frac{x+2}{\sqrt{(x+2)(x-2)}}=\frac{x+2}{\sqrt{x^2-4}}.

Now split the numerator:

∫x+2x2−4 dx=∫xx2−4 dx⏟I1+2∫dxx2−4⏟I2.\int\frac{x+2}{\sqrt{x^2-4}}\,dx=\underbrace{\int\frac{x}{\sqrt{x^2-4}}\,dx}_{I_1}+\underbrace{2\int\frac{dx}{\sqrt{x^2-4}}}_{I_2}.

I1I_1: substitute u=x2−4, du=2x dxu=x^2-4,\ du=2x\,dx:

I1=12∫u−1/2 du=u=x2−4.I_1=\frac12\int u^{-1/2}\,du=\sqrt{u}=\sqrt{x^2-4}.

I2I_2: standard result ∫dxx2−a2=ln⁡∣x+x2−a2∣\displaystyle\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right| with a=2a=2:

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

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