If int frac{x^2+1}{x^4+1} dx = f(x) + c,then | vedclass

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

Question

(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{1}{x^2}}{x^2 +

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight)$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$ We can rewrite the denominator as $(x - \frac{1}{x})^2 + 2$: $I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + (\sqrt{2})^2} dx$ Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{t}{\sqrt{2}}ight) + c$ Substituting $t = x - \frac{1}{x} = \frac{x^2-1}{x}$ back: $I = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight) + c$ Thus,$f(x) = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight)$.

If $\int \frac{x^2+1}{x^4+1} dx = f(x) + c$,then $f(x)$ is equal to

Similar Questions

$\int {{x^x}(1 + \log x)\,dx} $ is equal to

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

$\int \sqrt{5-2x+x^2} dx$ is equal to

If $\int \frac{x}{x \tan x+1} \, dx = \log f(x) + k$,then $f\left(\frac{\pi}{4}\right) =$

$\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module