int frac{x^2-1}{x^3 sqrt{2 x^4-2 x^2+1}} d x= | vedclass

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

Question

(D) We have, $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$. Divide the numerator and denominator inside the square root by $x^4$: $I = \int \fr

Vedclass · Accepted Answer

$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$

Vedclass · Answer

(D) We have, $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$. Divide the numerator and denominator inside the square root by $x^4$: $I = \int \frac{x^2-1}{x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{x^2-1}{x^5 \sqrt{2 - 2x^{-2} + x^{-4}}} dx$. $I = \int \frac{x^{-3} - x^{-5}}{\sqrt{2 - 2x^{-2} + x^{-4}}} dx$. Let $t = \sqrt{2 - 2x^{-2} + x^{-4}}$. Then $t^2 = 2 - 2x^{-2} + x^{-4}$. Differentiating both sides with respect to $x$: $2t \frac{dt}{dx} = 4x^{-3} - 4x^{-5} = 4(x^{-3} - x^{-5})$. So, $(x^{-3} - x^{-5}) dx = \frac{1}{2} t dt$. Substituting this into the integral: $I = \int \frac{\frac{1}{2} t dt}{t} = \frac{1}{2} \int dt = \frac{1}{2} t + c$. Substituting back $t$: $I = \frac{1}{2} \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + c$.

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$

Similar Questions

Evaluate $\int \sqrt{e^{4x} + e^{2x}} \, dx$.

$\int \frac{dx}{(a^2 + x^2)^{3/2}}$ is equal to

If $l^r(x)$ denotes the $r$-th iterated logarithm of $x$,i.e.,$l^1(x) = \log(x)$,$l^2(x) = \log(\log(x))$,...,$l^r(x) = \log(\log(...\log(x)...))$,then $\int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx = $

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module