The value of int frac{d x}{(x+1)^{3 / 4}(x-2) | vedclass

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

Question

(D) Let $I = \int \frac{d x}{(x+1)^{3/4} (x-2)^{5/4}}$.We can rewrite the integrand as:$I = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{

Vedclass · Accepted Answer

$\frac{-4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+c$,where $c$ is a constant of integration.

Vedclass · Answer

(D) Let $I = \int \frac{d x}{(x+1)^{3/4} (x-2)^{5/4}}$.We can rewrite the integrand as:$I = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{3/4} (x-2)^{5/4}} = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^2}$.Let $t = \frac{x+1}{x-2}$.Then,$dt = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} dx = \frac{-3}{(x-2)^2} dx$.Thus,$\frac{dx}{(x-2)^2} = -\frac{1}{3} dt$.Substituting these into the integral:$I = \int \frac{1}{t^{3/4}} \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int t^{-3/4} dt$.Integrating with respect to $t$:$I = -\frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + c = -\frac{4}{3} t^{1/4} + c$.Substituting back $t = \frac{x+1}{x-2}$:$I = -\frac{4}{3} \left(\frac{x+1}{x-2}\right)^{1/4} + c$.

The value of $\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to

Similar Questions

$\int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx = $

Evaluate the integral: $\int \frac{dx}{2+\cos x}$

If $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$,then $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$

If $\int \sqrt{\frac{x-7}{x-9}} ~dx = A \sqrt{x^2-16x+63} + \log \left|(x-8)+\sqrt{x^2-16x+63}\right| + c,$ (where $c$ is a constant of integration) then $A$ is

If the integral $\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} dx = A \cos 8x + k,$ where $k$ is an arbitrary constant,then $A$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module