Question

The value of the integral ${\int }_1^{3^{1/n}}\frac{dx}{x(x^n+1)}$ is

• A. $\frac{1}{n}\log \left(\frac{2}{3}\right)$
• B. $n\log \left(\frac{2}{3}\right)$
• C. $\frac{1}{n}\log \left(\frac{3}{2}\right)$
• D. $n\log \left(\frac{3}{2}\right)$

Answer 1

Answer: (C)

$\frac{1}{n}\log \left(\frac{3}{2}\right)$

Let $x^n=t$ $\Rightarrow$ $n{x}^{n-1}dx=dt$
$\Rightarrow$ $\frac{n{x}^n}{x}dx=dt$
$\Rightarrow$ $\frac{1}{x}dx=\frac{dt}{n{x}^n}$
$\Rightarrow$ $\frac{1}{x}dx=\frac{dt}{nt}$
$\therefore {\int }_1^{3^{1/n}}\frac{dx}{x(x^n+1)}={\int }_1^3\frac{1}{nt(t+1)}dt$
$=\frac{1}{n}{\int }_1^3\frac{1}{t(t+1)}dt=\frac{1}{n}{\int }_1^3(\frac{1}{t}-\frac{1}{t+1})dt=\frac{1}{n}{[\log t-\log (t+1)]}_1^3$
$=\frac{1}{n}[\log \left(\frac{3}{4}\right)-\log \left(\frac{1}{2}\right)]=\frac{1}{n}\left[\log (\frac{3}{4}\times \frac{2}{1})\right]$
$=\frac{1}{n}\log \left(\frac{3}{2}\right)$