A) \[-log\text{ }\left[ log\left( \frac{x+1}{x} \right) \right]+c\]
B) \[-log\text{ }log\left( \frac{x+1}{x} \right)+c\]
C) \[-{{\left\{ log\left( x+1 \right) \right\}}^{2}}-{{\left\{ log\left( x \right) \right\}}^{2}}+c\]
D) \[-\frac{1}{2}.{{\left[ \log \frac{(x+1)}{x} \right]}^{2}}+c\]
Correct Answer: D
Solution :
[d] \[\because I=\int{\frac{\log (1+x)-\log x}{x(1+x)}}.dx=\int{\frac{\log \left( \frac{x+1}{x} \right)}{x(1+x)}.dx}\]let log \[\Rightarrow \left( \frac{1+x}{x} \right)=z\] \[\Rightarrow \frac{1}{\frac{1+x}{x}}\times \frac{d}{dx}\left( \frac{1+x}{x} \right).dx=dz\] \[\Rightarrow \frac{x}{1+x}.\left\{ \frac{x(0+1)-(1+x).1}{{{x}^{2}}} \right\}.dx=dz\] \[\Rightarrow \frac{x}{1+x}\times \frac{-1}{{{x}^{2}}}.dx=dz\] \[\Rightarrow \frac{-dx}{x(1+x)}=dz\] Now, \[I=\int{z.dz}=-\frac{{{z}^{2}}}{z}+c=\frac{-1}{2}.{{\left\{ \log \left( \frac{1+x}{x} \right) \right\}}^{2}}+c\] Hence, option [d] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
