int frac{1}{x^2(x^4 + 1)^{3/4}} dx = | vedclass

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

Question

(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$. Taking $x^4$ common from the bracket,we get: $I = \int \frac{1}{x^2 [x^4(1 + \fra

Vedclass · Accepted Answer

$-\frac{(x^4 + 1)^{1/4}}{x} + c$

Vedclass · Answer

(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$. Taking $x^4$ common from the bracket,we get: $I = \int \frac{1}{x^2 [x^4(1 + \frac{1}{x^4})]^{3/4}} dx = \int \frac{1}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} dx = \int \frac{1}{x^5 (1 + \frac{1}{x^4})^{3/4}} dx$. 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: $I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$. Substituting $t$ back: $I = -(1 + \frac{1}{x^4})^{1/4} + c = -(\frac{x^4 + 1}{x^4})^{1/4} + c = -\frac{(x^4 + 1)^{1/4}}{x} + c$.

$\int \frac{1}{x^2(x^4 + 1)^{3/4}} dx = $

Similar Questions

If $f(x) = \int \frac{1}{x^{1/4}(1+x^{1/4})} dx$ and $f(0) = -6$,then $f(1)$ is equal to:

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k$ is equal to

$\int \tan ^8 x \cdot \sec ^4 x \, dx = $ . . . . . . $+ C$.

Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module