Question

$I=\int \frac{\log xdx}{{\left(1+\log x\right)}^2}$

• A. $e^{\tan x}(x^2-\sin x)+c$
• B. $e^{\tan x}(x+\sin 2x)+c$
• C. $e^{\tan x}(x-\cos 2x)+c$
• D. $\frac{x}{\left(\log x+1\right)}+C.$

Answer 1

The correct option is D $\frac{x}{\left(\log x+1\right)}+C.$

$\begin{array}{c}I=\int \frac{\log xdx}{{\left(1+\log x\right)}^2}\\ \log =t\\ x=e^t\\ dx=e^{t}dt\\ \int \frac{t{e}^t}{\left(1+t^2\right)}dt\\ =\int e^t\left(\frac{1}{t+1}-\frac{1}{{\left(t+1\right)}^2}\right)dt\\ =\frac{e^t}{t+1}+C\\ =\frac{x}{\left(\log x+1\right)}+C.\end{array}$

Hence, the option $D$ is the correct answer.