Question

Evaluate the following integrals:

\[\int\frac{\log x}{\left( x + 1 \right)^2}dx\]

 

Answer 1

\[\text{ Let I }= \int\frac{\log x}{\left( x + 1 \right)^2}dx\]
`  "Let the first function be ( log x ) and second function be   "1/(x+1)^2"   "   `
\[\text{First we find the integral of the second function, i . e} . , \int\frac{1}{\left( x + 1 \right)^2}dx . \]
\[\text{ Put t } = \left( x + 1 \right) . Then dt = dx\]
\[\text{ Therefore,} \]
\[\int\frac{1}{\left( x + 1 \right)^2}dx = \int t^{- 2} dt\]
\[ = - \frac{1}{t}\]
\[ = - \frac{1}{1 + x}\]
\[\text{Hence, using integration by parts, we get}\]
\[\int\frac{\log x}{\left( x + 1 \right)^2}dx = \left( \log x \right)\int\frac{1}{\left( x + 1 \right)^2}dx - \int\left[ \left( \frac{d \left( \log x  \right)}{d x} \right)\int\frac{1}{\left( x + 1 \right)^2}dx \right]dx\]
\[ = \left( \log x \right)\left( - \frac{1}{1 + x} \right) - \int\left( \frac{1}{x} \right)\left( - \frac{1}{1 + x} \right)dx\]
\[ = - \frac{\log x}{1 + x} + \int\left( \frac{1}{x^2 + x} \right)dx\]
\[ = - \frac{\log x}{1 + x} + \int\frac{1}{x^2 + x + \frac{1}{4} - \frac{1}{4}}dx\]
\[ = - \frac{\log x}{1 + x} + \int\frac{1}{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}dx\]
\[ = - \frac{\log x}{1 + x} + \frac{1}{2 \times \frac{1}{2}}\text{ log }\left| \frac{x + \frac{1}{2} - \frac{1}{2}}{x + \frac{1}{2} + \frac{1}{2}} \right| + c\]
\[ = - \frac{\log x}{1 + x} + \text{ log }\left| \frac{x}{x + 1} \right| + c\]
\[\text{ Hence,} \int\frac{\log x}{\left( x + 1 \right)^2}dx = - \frac{\log x}{1 + x} + \text{ log}\left| \frac{x}{x + 1} \right| + c\]