Evalute : x (1+ x)^2 d x - Vidyadip

Question

Evalute : $\int \frac{\log x}{(1+\log x)^2} d x$

✓

Answer

Let $I=\int \frac{\log x}{(1+\log x)^2} d x$
Put $\log x=t \quad \therefore x=e^t$
$
\therefore d x=e^t d t
$
$
\begin{aligned}
\therefore I & =\int \frac{t}{(1+t)^2} \cdot e^t d t \\
& =\int e^t\left[\frac{(1+t)-1}{(1+t)^2}\right] d t \\
& =\int e^t\left[\frac{1}{1+t}-\frac{1}{(1+t)^2}\right] d t
\end{aligned}
$
Let $f(t)=\frac{1}{1+t}$.
$
\begin{aligned}
\therefore f^{\prime}(t) & =\frac{d}{d t}(1+t)^{-1}=-1(1+t)^{-2}(0+1) \\
& =\frac{-1}{(1+t)^2}
\end{aligned}
$
$
\begin{aligned}
\therefore I & =\int e^t\left[f(t)+f^{\prime}(t)\right] d t \\
& =e^t \cdot f(t)+c=e^t \times \frac{1}{1+t}+c=\frac{x}{1+\log x}+c .
\end{aligned}
$

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