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Question: \(\int_{}^{}\frac{\mathbf{(x}\mathbf{-}\mathbf{1)dx}}{\mathbf{(x + 1)}\sqrt{\mathbf{x(}\mathbf{x}^{\...

∫(x−1)dx(x+1)x(x2+x+1)\int_{}^{}\frac{\mathbf{(x}\mathbf{-}\mathbf{1)dx}}{\mathbf{(x + 1)}\sqrt{\mathbf{x(}\mathbf{x}^{\mathbf{2}}\mathbf{+ x + 1)}}}is

A

tan⁡−1x+1x+1\mathbf{\tan}^{\mathbf{- 1}}\sqrt{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}}+ c

B

2tan⁡−1x+1x+1\sqrt{\mathbf{2}}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}}+ l

C

2tan⁡–1x+1x+1\mathbf{2}\mathbf{\tan}^{\mathbf{–1}}\sqrt{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}}+ c

D

12tan⁡−1x+1x+1\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}}+ c

Answer

2tan⁡–1x+1x+1\mathbf{2}\mathbf{\tan}^{\mathbf{–1}}\sqrt{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}}+ c

Explanation

Solution

I = ∫(x−1)dx(x+1)xx+1x+1\int \frac { ( x - 1 ) d x } { ( x + 1 ) x \sqrt { x + \frac { 1 } { x } + 1 } }

= ∫(x2−1)dx(x2+2x+1)xx+1x+1\int \frac { \left( x ^ { 2 } - 1 \right) d x } { \left( x ^ { 2 } + 2 x + 1 \right) x \sqrt { x + \frac { 1 } { x } + 1 } }

=

put x + 1/x + 1 = t2