Question

Evaluate the integral
${\int }_1^{\sqrt[7]{72}}\frac{1}{x(2x^7+1)}dx$

• A. $\log \frac{6}{5}$
• B. $6\log \frac{6}{5}$
• C. $\frac{1}{7}\log \frac{6}{5}$
• D. $\frac{1}{5}\log \frac{6}{5}$

Answer 1

Answer: (C)

$\frac{1}{7}\log \frac{6}{5}$

${\int }_1^{\sqrt[7]{72}}\frac{1}{x(2x^7+1)}dx={\int }_1^{\sqrt[7]{72}}\frac{1}{x^8(2+1/x^7)dx}$

$1/x^7=1\Rightarrow \frac{-7}{x^8}dx=dt$

$\Rightarrow \frac{dx}{x^8}-\frac{dx}{7}$

$\Rightarrow \frac{1}{-7}{\int }_1^{1/2}\frac{dt}{(2+t)}=\frac{1}{7}{\int }_{1/2}^1{\left[\log \right(2+t\left)\right]}_{1/2}^1$

$=\frac{1}{7}\left[\log \right(3)-\log \frac{5}{2}]$

$=\frac{1}{7}\log \frac{6}{5}$