Question
Evaluate the integral: $\int \frac{1}{x \sqrt{1+x^n}} d x$
✓
Answer
Let the given integral be,
$I=\int \frac{d x}{x \sqrt{1+x^n}}$
$=\int \frac{x^{n-1} d x}{x^{n-1} x^1 \sqrt{1+x^n}}$
$=\int \frac{x^{n-1} d x}{x^n \sqrt{1+x^n}}$
Putting $x ^{ n }= t$
$\Rightarrow nx x^{n-1} dx=dt
\\\Rightarrow x^{n-1} dx=\frac{d t}{n}
\\\therefore I=\frac{1}{n} \int \frac{d t}{t \sqrt{1+t}}$
let $1+ t = p ^2$
$\Rightarrow dp=2 p dp$
$\therefore I=\frac{1}{n} \int \frac{2 pdp}{\left(p^2-1\right) p}$
$=\frac{2}{n} \int \frac{d p}{p^2-1^2}$
$=\frac{2}{n} \times \frac{1}{2} \log \left|\frac{p-1}{p+1}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+t-1}}{\sqrt{1+t+1}}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+C$
$I=\int \frac{d x}{x \sqrt{1+x^n}}$
$=\int \frac{x^{n-1} d x}{x^{n-1} x^1 \sqrt{1+x^n}}$
$=\int \frac{x^{n-1} d x}{x^n \sqrt{1+x^n}}$
Putting $x ^{ n }= t$
$\Rightarrow nx x^{n-1} dx=dt
\\\Rightarrow x^{n-1} dx=\frac{d t}{n}
\\\therefore I=\frac{1}{n} \int \frac{d t}{t \sqrt{1+t}}$
let $1+ t = p ^2$
$\Rightarrow dp=2 p dp$
$\therefore I=\frac{1}{n} \int \frac{2 pdp}{\left(p^2-1\right) p}$
$=\frac{2}{n} \int \frac{d p}{p^2-1^2}$
$=\frac{2}{n} \times \frac{1}{2} \log \left|\frac{p-1}{p+1}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+t-1}}{\sqrt{1+t+1}}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+C$
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