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