Indefinite IntegrationmediumFree

Integral of x/(2−x²+√(2−x²)) | JEE

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

Question

\displaystyle\int \dfrac{x}{2-x^{2}+\sqrt{2-x^{2}}}\,dx

equals

-x\ln\big|1-\sqrt{2-x^{2}}\big|+C

\ln\big|1+\sqrt{2+x^{2}}\big|+C

-\ln\big|1+\sqrt{2-x^{2}}\big|+C

correct

x\ln\big|1-\sqrt{2+x^{2}}\big|+C

Solution

Step 1: Substitute

x=\sqrt{2}\sin\theta

, so

dx=\sqrt{2}\cos\theta\,d\theta

and

2-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-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=\sqrt{2}\sin\theta\cdot\sqrt{2}\cos\theta\,d\theta=2\sin\theta\cos\theta\,d\theta

, so

I=\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=\sqrt{2}\cos\theta+1\Rightarrow du=-\sqrt{2}\sin\theta\,d\theta

I=\int\dfrac{-du}{u}=-\ln|u|+C=-\ln\big|\sqrt{2}\cos\theta+1\big|+C

Step 5: Back-substitute

\sqrt{2}\cos\theta=\sqrt{2-x^{2}}

I=-\ln\big|1+\sqrt{2-x^{2}}\big|+C

Correct answer: (3)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.