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

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

Question

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$.We can rewrite the integral as $I = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{\frac{3}{

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

$\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \,d x=$

$\int \left( \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} \right) dx =$

If $\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$,then $\frac{P}{Q} =$

$ \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x $

Find $\int \cos 6x \sqrt{1+\sin 6x} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module