Question

The value of the integral $\displaystyle \int \frac{d x}{\left(x - 2\right)^{\frac{7}{8}} \left(x + 3\right)^{\frac{9}{8}}}$ 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 $I=\int \frac{d x}{(x-2)^{7 / 8}(x+3)^{9 / 8}}$
On substituting $\frac{x-2}{x+3}=t$, we get
$\frac{(x+3) \times 1-(x-2) \times 1}{(x+3)^{2}} d x=d t$
$
\Rightarrow \frac{d x}{(x+3)^{2}}=\frac{d t}{5}
$
$
\therefore I=\int \frac{(x+3)^{7 / 8} d x}{(x-2)^{7 / 8}(x+3)^{7 / 8}(x+3)^{9 / 8}}
$
$
=\int\left(\frac{x+3}{x-2}\right)^{7 / 8} \frac{d x}{(x+3)^{2}}
$
$
\begin{array}{l}
=\int\left(\frac{1}{t}\right)^{7 / 8} \frac{d t}{5}=\frac{1}{5} \int t^{-7 / 8} d t \\
=\frac{1}{5}\left[\frac{t^{1 / 8}}{1 / 8}\right]+C \\
=\frac{8}{5}\left(\frac{x-2}{x+3}\right)^{1 / 8}+C
\end{array}
$