Integrate the function x (1+ x)^2 with respect to x. - Vidyadip

Question

Integrate the function $\frac{\log x}{(1+\log x)^2}$ with respect to $x$.

✓

Answer

Let $\quad I =\int \frac{\log x}{(1+\log x)^2} d x$
Adding and subtracting 1 from numerator
$
\begin{array}{l}
I=\int \frac{(\log x+1-1)}{(1+\log x)^2} d x \\
I=\int \frac{1}{(1+\log x)} d x-\int \frac{d x}{(1+\log x)^2} \\
I=\int \frac{1}{(1+\log x)} \cdot 1 d x-\int \frac{d x}{(1+\log x)^2}
\end{array}
$
Assuming " 1 " as second function integrating by parts :
$
\begin{array}{l}
I=\frac{1}{(1+\log x)} \cdot x-\int \frac{-1}{(1+\log x)^2} \cdot \frac{1}{x} \times x d x- \\
\int \frac{d x}{(1+\log x)^2}+C \\
I=\frac{x}{(1+\log x)}+\int \frac{1}{(1+\log x)^2} d x- \\
\int \frac{1}{(1+\log x)^2} d x+C \\
I=\frac{x}{(1+\log x)}+C \text { Ans. }
\end{array}
$

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