Question

The value of $\int \frac{\left(x-2\right)dx}{{\left\{{\left(x-2\right)}^2{\left(x+3\right)}^7\right\}}^{1/3}}dx$ is

• A. $\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{4/3}+C$
• B. $\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{3/4}+C$
• C. $\frac{5}{12}{\left(\frac{x-2}{x+3}\right)}^{4/3}+C$
• D. $\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{5/3}+C$

Answer 1

Answer: (A)

Let $I=\int \frac{(x-2)}{{\left\{(x-2{)}^2(x+3{)}^7\right\}}^{1/3}}dx$
$=\int \frac{dx}{(x-2{)}^{-1/3}(x+3{)}^{7/3}}$
$=\int \frac{dx}{(x-2{)}^2{\left(\frac{x+3}{x-2}\right)}^{7/3}}$
Put $\frac{x+3}{x-2}=t$
$\Rightarrow \frac{-5dx}{(x-2{)}^2}=dt$
$\therefore I=-\frac{1}{5}\int \frac{dt}{t^{7/3}}$
$=-\frac{1}{5}\int t^{-7/3}dt$
$=-\frac{1}{5}\left[\frac{t^{-7/3+1}}{-\frac{7}{3}+1}\right]+C$
$=-\frac{1}{5}\left[\frac{t^{-4/3}}{-\frac{4}{3}}\right]+C$
$=\frac{3}{20t^{4/3}}+C$
$=\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{4/3}+C$