Question

$\int_{}^{}{\frac{e^{2\tan^{- 1}x}(1 + x)^{2}}{1 + x^{2}}dx =}$

• A. $xe^{\tan^{- 1}x} + c$
• B. $xe^{2\tan^{- 1}x} + c$
• C. $2xe^{\tan^{- 1}x} + c$
• D. None of these

Answer 1

Answer: (B)

$xe^{2\tan^{- 1}x} + c$

Answer 2

Put $\tan^{- 1}x = t$ $\Rightarrow \frac{1}{1 + x^{2}}dx = dt$

$\therefore I = \int_{}^{}\frac{e^{2\tan^{- 1}x}(1 + x)^{2}}{1 + x^{2}}dx = \int_{}^{}{e^{2t}(1 + \tan t)^{2}dt}$=$\int_{}^{}{e^{2t}(\sec^{2}t + 2\tan t)dt}$ $I = \int_{}^{}{e^{2t}\sec^{2}tdt + 2\left\lbrack \tan t\frac{e^{2t}}{2} - \int_{}^{}{\sec^{2}t.\frac{e^{2t}}{2}dt} \right\rbrack}$

$\Rightarrow I = \int_{}^{}{e^{2t}\sec^{2}tdt + e^{2t}\tan t - \int_{}^{}{e^{2t}\sec^{2}tdt + c}}I = e^{2t}\tan t + c$ $\Rightarrow$ $I = xe^{2\tan^{- 1}x} + c$