Question

The value of the integral $\int \frac{dx}{{\left(x-2\right)}^{\frac{7}{8}}{\left(x+3\right)}^{\frac{9}{8}}}$ is equal to (where, $C$ is the constant of integration)

• A. $\frac{8}{5}{\left(\frac{x-2}{x+3}\right)}^{\frac{1}{8}}+C$
• B. $\frac{5}{8}{\left(\frac{x-2}{x+3}\right)}^{\frac{1}{8}}+C$
• C. $\frac{5}{8}{\left(\frac{x+3}{x-2}\right)}^{\frac{1}{8}}+C$
• D. $\frac{8}{5}{\left(\frac{x+3}{x-2}\right)}^{\frac{1}{8}}+C$

Answer 1

Answer: (A)

Let $I=\int \frac{dx}{(x-2{)}^{7/8}(x+3{)}^{9/8}}$
On substituting $\frac{x-2}{x+3}=t$, we get
$\frac{(x+3)\times 1-(x-2)\times 1}{(x+3{)}^2}dx=dt$
$<br/>\Rightarrow \frac{dx}{(x+3{)}^2}=\frac{dt}{5}<br/>$
$<br/>\therefore I=\int \frac{(x+3{)}^{7/8}dx}{(x-2{)}^{7/8}(x+3{)}^{7/8}(x+3{)}^{9/8}}<br/>$
$<br/>=\int {\left(\frac{x+3}{x-2}\right)}^{7/8}\frac{dx}{(x+3{)}^2}<br/>$
$<br/>\begin{array}{l}<br/>=\int {\left(\frac{1}{t}\right)}^{7/8}\frac{dt}{5}=\frac{1}{5}\int t^{-7/8}dt\\ <br/>=\frac{1}{5}\left[\frac{t^{1/8}}{1/8}\right]+C\\ <br/>=\frac{8}{5}{\left(\frac{x-2}{x+3}\right)}^{1/8}+C<br/>\end{array}<br/>$