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Question

Mathematics Question on Integrals of Some Particular Functions

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

A

11+xlogx+C\frac{1}{1+x\,log\,\left|x\right|}+C

B

11+logx+C\frac{1}{1+log\,\left|x\right|}+C

C

11+xlogx+C\frac{-1}{1+x\,log\,\left|x\right|}+C

D

log11+logx+Clog\left|\frac{1}{1+log\,\left|x\right|}\right|+C

Answer

11+xlogx+C\frac{-1}{1+x\,log\,\left|x\right|}+C

Explanation

Solution

Let I=1+logx(1+xlogx)2dxI=\int \frac{1+\log x}{(1+x \log x)^{2}} d x Let 1+xlogx=t1+x \log x=t (0+x×1x+logx)dx=dt\Rightarrow\left(0+x \times \frac{1}{x}+\log x\right) d x=d t (1+logx)dx=dt\Rightarrow (1+\log x) d x=d t I=1t2dt=t11+C\therefore I=\int \frac{1}{t^{2}} d t=\frac{t^{-1}}{-1}+C =1(1+xlogx)+C=\frac{-1}{(1+x \log x)}+C