Question

The value of the integral $\int (x^2+x)(x^{-8}+2x^{-9}{)}^{\frac{1}{10}}dx$ is

• A. $\frac{5}{11}(x^2+2x{)}^{\frac{11}{10}}+c$
• B. $\frac{5}{6}(x+1{)}^{\frac{11}{10}}+c$
• C. $\frac{6}{7}(x+1{)}^{\frac{11}{10}}+c$
• D. none of these

Answer 1

The correct option is A $\frac{5}{11}(x^2+2x{)}^{\frac{11}{10}}+c$

Given that $I=\int (x^2+x){(x^{-8}+2x^{-9})}^{\frac{1}{10}}dx$
$=\int x(x+1){\left(\frac{x+2}{x^9}\right)}^{\frac{1}{10}}dx$
$=\int (x+1)(x^2+2x{)}^{\frac{1}{10}}dx$
Substitute $x^2+2x=t\Rightarrow (x+1)dx=\frac{dt}{2}$
$\therefore I=\int t^{\frac{1}{10}}\frac{dt}{2}$
$=\frac{1}{2}\times \frac{10}{11}{t}^{\frac{11}{10}}$
$I=\frac{5}{11}(x^2+2x{)}^{\frac{11}{10}}+C$