Question

Evaluate $\int \frac{dx}{(x-2{)}^{7/8}(x+3{)}^{9/8}}$

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

Answer 1

Answer: (A)

$\frac{8}{5}{\left(\frac{x-2}{x+3}\right)}^{1/8}+c$

$I=\int \frac{dx}{(x-2{)}^{7/8}(x+3{)}^{9/8}}$
$I=\int \frac{dx}{{\left(\frac{x-2}{x+3}\right)}^{7/8}(x+3{)}^2}$
Put $\frac{x-2}{x+3}=t$
i.e. $\frac{5}{(x+3{)}^2}dx=dt$
So, $I=\frac{1}{5}\int \frac{1}{t^{7/8}}dt$
$I=\frac{8}{5}{t}^{1/8}+C$
$\therefore I=\frac{8}{5}{\left(\frac{x-2}{x+3}\right)}^{1/8}+C$