First slide

Methods of integration

Question

The integral $\int \frac{dx}{(x + 4)^{8 / 7} (x - 3)^{6 / 7}}$ is equal to (where C is a constant of integration)

Moderate

A

$- {(\frac{x - 3}{x + 4})}^{- 1 / 7} + C$
B

$\frac{1}{2} {(\frac{x - 3}{x + 4})}^{3 / 7} + C$
C

${(\frac{x - 3}{x + 4})}^{1 / 7} + C$
D

$- \frac{1}{13} {(\frac{x - 3}{x + 4})}^{- 13 / 7} + C$

Solution

The given integral,
$\begin{array}{l} I = \int \frac{dx}{(x + 4)^{8 / 7} (x - 3)^{6 / 7}} \\ = \int \frac{\frac{dx}{(x + 4)^{2}}}{\frac{(x + 4)^{8 / 7} (x - 3)^{6 / 7}}{(x + 4)^{2}}} = \int \frac{\frac{dx}{(x + 4)^{2}}}{{(\frac{x - 3}{x + 4})}^{6 / 7}} \end{array}$
Now, let $\frac{x - 3}{x + 4} = t^{7}$
$\begin{array}{l} \Rightarrow \frac{(x + 4) - (x - 3)}{(x + 4)^{2}} dx = 7 t^{6} dt \\ \Rightarrow \frac{7}{(x + 4)^{2}} dx = 7 t^{6} dt \end{array}$
So, $I = \int \frac{t^{6} dt}{t^{6}} = \int dt = t + C$
$∴ I = {(\frac{x - 3}{x + 4})}^{1 / 7} + C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App