int frac{x^4+1}{x^6+1} dx = | vedclass

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

Question

(B) Step $1$: Rewrite the numerator as $(x^4 - x^2 + 1) + x^2$. Step $2$: Split the integral into two parts: $I = \int \frac{x^4 - x^2 + 1}{x^6 + 1} d

Vedclass · Accepted Answer

$	an^{-1} x + \frac{1}{3} 	an^{-1}(x^3) + c$

Vedclass · Answer

(B) Step $1$: Rewrite the numerator as $(x^4 - x^2 + 1) + x^2$. Step $2$: Split the integral into two parts: $I = \int \frac{x^4 - x^2 + 1}{x^6 + 1} dx + \int \frac{x^2}{x^6 + 1} dx$. Step $3$: Note that $x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$. Thus, the first integral becomes $\int \frac{1}{x^2 + 1} dx = 	an^{-1} x$. Step $4$: For the second integral, let $u = x^3$, then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$. The integral becomes $\int \frac{1}{3(u^2 + 1)} du = \frac{1}{3} 	an^{-1}(u) = \frac{1}{3} 	an^{-1}(x^3)$. Step $5$: Combining both, we get $	an^{-1} x + \frac{1}{3} 	an^{-1}(x^3) + c$.

$\int \frac{x^4+1}{x^6+1} dx = $

Similar Questions

$\int \frac{3 \sin x+5 \cos x+4}{\sin x+\cos x+2} d x=$

$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ is equal to

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

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

$\int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module