Evalute : 1+ x x(3+ x)(2+3 x) d x - Vidyadip

Question

Evalute : $\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x$

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Answer

Let $I=\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x$
$
=\int \frac{1+\log x}{(3+\log x)(2+3 \log x)} \cdot \frac{1}{x} d x
$
Put $\log x=t \quad \therefore \frac{1}{x} d x=d t$
$
\therefore I=\int \frac{1+t}{(3+t)(2+3 t)} d t
$
Let $\frac{1+t}{(3+t)(2+3 t)}=\frac{A}{3+t}+\frac{B}{2+3 t}$
$
\therefore 1+t=A(2+3 t)+B(3+t)
$
Put $3+t=0$, i.e. $t=-3$, we get
$
1-3=A(-7)+B(0)
$
$
\therefore-2=-7 A \quad \therefore A=\frac{2}{7}
$
Put $2+3 t=0$, i.e. $t=-\frac{2}{3}$, we get
$
1-\frac{2}{3}=A(0)+B\left(\frac{7}{3}\right)
$
$
\therefore \frac{1}{3}=\frac{7}{3} B \quad \therefore B=\frac{1}{7}
$
$
\therefore \frac{1+t}{(3+t)(2+3 t)}=\frac{\left(\frac{2}{7}\right)}{3+t}+\frac{\left(\frac{1}{7}\right)}{2+3 t}
$
$\begin{aligned} \therefore I & =\int\left[\frac{\left(\frac{2}{7}\right)}{3+t}+\frac{\left(\frac{1}{7}\right)}{2+3 t}\right] d t \\ & =\frac{2}{7} \int \frac{1}{3+t} d t+\frac{1}{7} \int \frac{1}{2+3 t} d t \\ & =\frac{2}{7} \log |3+t|+\frac{1}{7} \cdot \frac{\log |2+3 t|}{3}+c \\ & =\frac{2}{7} \log |3+\log x|+\frac{1}{21} \log |2+3 \log x|+c .\end{aligned}$

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