Question

\int\left\\{\frac{\left(log \,x -1\right)}{1+\left(log \,x\right)^{2}}\right\\}^{2}dx is equal to :

• A. $\frac{x}{\left(log\,x\right)^{2}+1}+c$
• B. $\frac{xe^{x}}{1+x^{2}}+c$
• C. $\frac{x}{x^{2} +1}+c$
• D. $\frac{log\,x}{\left(log\,x\right)^{2} +1}+c$

Answer 1

Answer: (A)

$\frac{x}{\left(log\,x\right)^{2}+1}+c$

Answer 2

Let $I = \int \left(\frac{log\,x-1}{ 1+\left(log\,x\right)^{2}}\right)^{2} dx$ $= \int \frac{\left(log\,x\right)^{2}+1-2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$ $= \int \frac{dx}{1+\left(log\,x\right)^{2}}- \int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$ $= \frac{x}{1+\left(log\,x\right)^{2}}+\int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$ $-\int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}$ $= \frac{x}{1+\left(log\,x\right)^{2}}+c$