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Integral of x/(2−x²+√(2−x²)) | JEE

JEE Maths question with a full step-by-step solution.

Question
∫x2−x2+2−x2 dx\displaystyle\int \dfrac{x}{2-x^{2}+\sqrt{2-x^{2}}}\,dx equals
A−xln⁡∣1−2−x2∣+C-x\ln\big|1-\sqrt{2-x^{2}}\big|+C
Bln⁡∣1+2+x2∣+C\ln\big|1+\sqrt{2+x^{2}}\big|+C
C−ln⁡∣1+2−x2∣+C-\ln\big|1+\sqrt{2-x^{2}}\big|+Ccorrect
Dxln⁡∣1−2+x2∣+Cx\ln\big|1-\sqrt{2+x^{2}}\big|+C
Solution
Step 1: Substitute x=2sin⁡θx=\sqrt{2}\sin\theta, so dx=2cos⁡θ dθdx=\sqrt{2}\cos\theta\,d\theta and
2−x2=2−2sin⁡2θ=2cos⁡2θ  ⇒  2−x2=2cos⁡θ2-x^{2}=2-2\sin^{2}\theta=2\cos^{2}\theta\;\Rightarrow\;\sqrt{2-x^{2}}=\sqrt{2}\cos\theta
 Step 2: Rewrite the denominator:
2−x2+2−x2=2cos⁡2θ+2cos⁡θ=2cos⁡θ(2cos⁡θ+1)2-x^{2}+\sqrt{2-x^{2}}=2\cos^{2}\theta+\sqrt{2}\cos\theta=\sqrt{2}\cos\theta\big(\sqrt{2}\cos\theta+1\big)
 Step 3: Form the numerator x dx=2sin⁡θ⋅2cos⁡θ dθ=2sin⁡θcos⁡θ dθx\,dx=\sqrt{2}\sin\theta\cdot\sqrt{2}\cos\theta\,d\theta=2\sin\theta\cos\theta\,d\theta, so
I=∫2sin⁡θcos⁡θ2cos⁡θ(2cos⁡θ+1) dθ=∫2sin⁡θ2cos⁡θ+1 dθI=\int\dfrac{2\sin\theta\cos\theta}{\sqrt{2}\cos\theta(\sqrt{2}\cos\theta+1)}\,d\theta=\int\dfrac{\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+1}\,d\theta
 Step 4: Substitute u=2cos⁡θ+1⇒du=−2sin⁡θ dθu=\sqrt{2}\cos\theta+1\Rightarrow du=-\sqrt{2}\sin\theta\,d\theta:
I=∫−duu=−ln⁡∣u∣+C=−ln⁡∣2cos⁡θ+1∣+CI=\int\dfrac{-du}{u}=-\ln|u|+C=-\ln\big|\sqrt{2}\cos\theta+1\big|+C
 Step 5: Back-substitute 2cos⁡θ=2−x2\sqrt{2}\cos\theta=\sqrt{2-x^{2}}:
I=−ln⁡∣1+2−x2∣+CI=-\ln\big|1+\sqrt{2-x^{2}}\big|+C
 Correct answer: (3)
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