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∫(tan⁻¹(x/(x²+1))+tan⁻¹((x²+1)/x)) dx | JEE

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

Question

The value of

\displaystyle\int_{-1}^{3}\left(\tan^{-1}\dfrac{x}{x^{2}+1}+\tan^{-1}\dfrac{x^{2}+1}{x}\right)dx

\pi

correct

2\pi

4\pi

DNone of these

Solution

Step 1: Use the complementary identity

\tan^{-1}a+\tan^{-1}\dfrac1a=\dfrac{\pi}{2}\operatorname{sgn}(a)

, valid for

a\ne 0

. Here

a=\dfrac{x}{x^{2}+1}

and

\dfrac1a=\dfrac{x^{2}+1}{x}

, so the integrand equals

\dfrac{\pi}{2}\operatorname{sgn}\!\left(\dfrac{x}{x^{2}+1}\right)

. Step 2: Determine the sign. Since

x^{2}+1>0

, the sign of

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

is just the sign of

x

. Hence the integrand is

\begin{cases}\dfrac{\pi}{2},&x>0,\\[4pt]-\dfrac{\pi}{2},&x<0.\end{cases}

Step 3: Split the integral at

x=0

\int_{-1}^{3}\dfrac{\pi}{2}\operatorname{sgn}\!\left(\dfrac{x}{x^{2}+1}\right)dx=\int_{-1}^{0}\!\left(-\dfrac{\pi}{2}\right)dx+\int_{0}^{3}\dfrac{\pi}{2}\,dx.

Step 4: Evaluate:

=-\dfrac{\pi}{2}(0-(-1))+\dfrac{\pi}{2}(3-0)=-\dfrac{\pi}{2}+\dfrac{3\pi}{2}=\pi.

Correct answer: (1)

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