Question
What is $$\int {\frac{{\log \,x}}{{{{\left( {1 + \log \,x} \right)}^2}}}dx} $$ equal to ?
Where $$c$$ is a constant
A.
$$\frac{1}{{{{\left( {1 + \log \,x} \right)}^3}}} + c$$
B.
$$\frac{1}{{{{\left( {1 + \log \,x} \right)}^2}}} + c$$
C.
$$\frac{x}{{\left( {1 + \log \,x} \right)}} + c$$
D.
$$\frac{x}{{{{\left( {1 + \log \,x} \right)}^2}}} + c$$
Answer :
$$\frac{x}{{\left( {1 + \log \,x} \right)}} + c$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int {\frac{{\log \,x}}{{{{\left( {1 + \log \,x} \right)}^2}}}dx} \cr
& {\text{Put }}\log \,x = t \Rightarrow \frac{1}{x}dx = dt \cr
& I = \int {\frac{{{e^t}t}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr
& = \int {\frac{{{e^t}.\left( {t + 1 - 1} \right)}}{{{{\left( {1 + t} \right)}^2}}}} dt \cr
& = \int {\frac{{{e^t}\left( {1 + t} \right)}}{{{{\left( {1 + t} \right)}^2}}}} dt - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr
& = \int {\frac{{{e^t}}}{{1 + t}}dt - } \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr
& = \frac{{{e^t}}}{{1 + t}} - \int { - {e^t}\frac{1}{{{{\left( {1 + t} \right)}^2}}}dt} - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr
& = \frac{{{e^t}}}{{1 + t}} + \int {^{{e^t}}\frac{1}{{{{\left( {1 + t} \right)}^2}}}dt} - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr
& = \frac{{{e^t}}}{{1 + t}} + c \cr
& = \frac{x}{{1 + \log \,x}} + c \cr} $$