The value of int frac{(x-2)}{{(x-2)^{2}(x+3)^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$.We can rewrite the integrand as:$I = \int \frac{(x-2)}{(x-2)^{2/3}(x+3)^{7/3}} d x

Vedclass · Accepted Answer

$\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{4 / 3}+C$

Vedclass · Answer

(A) Let $I = \int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$.We can rewrite the integrand as:$I = \int \frac{(x-2)}{(x-2)^{2/3}(x+3)^{7/3}} d x = \int \frac{(x-2)^{1/3}}{(x+3)^{7/3}} d x$.This can be written as:$I = \int \frac{1}{(x+3)^{7/3} \cdot (x-2)^{-1/3}} d x = \int \frac{1}{(x-2)^2 \cdot \left(\frac{x+3}{x-2}\right)^{7/3}} d x$.Let $t = \frac{x+3}{x-2}$. Then $dt = \frac{(x-2)(1) - (x+3)(1)}{(x-2)^2} dx = \frac{-5}{(x-2)^2} dx$.Thus, $\frac{dx}{(x-2)^2} = -\frac{1}{5} dt$.Substituting these into the integral:$I = -\frac{1}{5} \int t^{-7/3} dt = -\frac{1}{5} \left[ \frac{t^{-4/3}}{-4/3} \right] + C$.$I = -\frac{1}{5} \cdot \left( -\frac{3}{4} \right) t^{-4/3} + C = \frac{3}{20} t^{-4/3} + C$.Substituting $t = \frac{x+3}{x-2}$ back:$I = \frac{3}{20} \left( \frac{x+3}{x-2} \right)^{-4/3} + C = \frac{3}{20} \left( \frac{x-2}{x+3} \right)^{4/3} + C$.

The value of $\int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$ is

Similar Questions

$\int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx = $

The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to: (Here $C$ is a constant of integration)

If $\int \left( \frac{4 e^x - 25}{2 e^x - 5} \right) dx = Ax + B \log |2 e^x - 5| + C$,then:

$\int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx =$

Integrate the function $\frac{x}{9-4x^{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module