Question

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

The correct option is A $\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{4/3}+C$

Let $I=\int \frac{(x-2)}{\left\{\right(x-2{)}^2(x+3{)}^7{\}}^{1/3}}dx$

$=\int \frac{dx}{(x-2{)}^{-1/3}(x+3{)}^{7/3}}$

$=\int \frac{dx}{(x-2{)}^2{\left(\frac{x+3}{x-2}\right)}^{7/3}}$

Put $\frac{x+3}{x-2}=t\Rightarrow \frac{-5dx}{(x-2{)}^2}=dt$

$\therefore I=-\frac{1}{5}\int \frac{dt}{t^{7/3}}=-\frac{1}{5}{\int }^{-7/3}dt$

$=-\frac{1}{5}\left[\frac{t^{-7/3+1}}{-\frac{7}{3}+1}\right]+C=-\frac{1}{5}\left[\frac{t^{-4/3}}{-\frac{4}{3}}\right]+C$

$=\frac{3}{20t^{4/3}}+C=\frac{3}{20}{\left(\frac{x-2}{x+3}\right)}^{4/3}+C$