∫0∞tan−1(x)(1+x)xdx=

π 2 4 , ∫ 0 ∞ arctan ( 1 / x ) x ( 1 + x ) dx , π 2 ∫ 0 ∞ d x x ( 1 + x )

∫0∞tan−1(x)(1+x)xdx=

I = ∫ 0 ∞ arctan ( 1 / x ) x ( 1 + x ) dx = 1 2 ∫ 0 ∞ arctan ( x ) + arctan ( 1 / x ) x ( 1 + x ) d x = π 4 ∫ 0 ∞ d x x ( 1 + x ) = π 2 4

∫ 0 ∞ arctan ( 1 / x ) x ( 1 + x ) dx

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$\int_{0}^{\infty} \frac{\tan^{- 1} (x)}{(1 + x) \sqrt{x}} d x =$

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$\frac{π^{2}}{4}$

$\int_{0}^{\infty} \frac{\arctan (1 / x)}{\sqrt{x} (1 + x)} dx$

$\frac{π}{4} \int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)}$

$\frac{π}{2} \int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)}$

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{\infty} \frac{\arctan (1 / x)}{\sqrt{x} (1 + x)} dx \\ = \frac{1}{2} \int_{0}^{\infty} \frac{\arctan (x) + \arctan (1 / x)}{\sqrt{x} (1 + x)} d x \\ = \frac{π}{4} \int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)} = \frac{π^{2}}{4} \end{array}$

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