Question
Integrate the following functions w.r.t. x:
$\frac{1+x}{x-\sin (x+\log x)}$
$\frac{1+x}{x-\sin (x+\log x)}$
✓
Answer
$ \text { Let } I=\int \frac{1+x}{x \cdot \sin (x+\log x)} d x$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1+x}{x}\right) d x$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1}{x}+1\right) d x$
Put $x+\log x=t \quad \therefore\left(1+\frac{1}{x}\right) d x=d t$
$\therefore I=\int \frac{1}{\sin t} d t=\int \operatorname{cosec} t d t$
$=\log |\operatorname{cosec} t-\cot t|+c$
$=\log |\operatorname{cosec}(x+\log x)-\cot (x+\log x)|+c .$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1+x}{x}\right) d x$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1}{x}+1\right) d x$
Put $x+\log x=t \quad \therefore\left(1+\frac{1}{x}\right) d x=d t$
$\therefore I=\int \frac{1}{\sin t} d t=\int \operatorname{cosec} t d t$
$=\log |\operatorname{cosec} t-\cot t|+c$
$=\log |\operatorname{cosec}(x+\log x)-\cot (x+\log x)|+c .$
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