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

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

Question

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$. Divide the numerator and denominator by $x^{2}$: $I = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$. Divide the numerator and denominator by $x^{2}$: $I = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx$. Rewrite the denominator as $(x^{2} + \frac{1}{x^{2}} - 2) + 3 = (x - \frac{1}{x})^{2} + (\sqrt{3})^{2}$. So,$I = \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x})^{2} + (\sqrt{3})^{2}} dx$. Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^{2}}) dx$. Substituting these into the integral,we get: $I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$. Using the standard integral formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = \frac{1}{\sqrt{3}} 	an^{-1}(\frac{t}{\sqrt{3}}) + c$. Replacing $t$ with $x - \frac{1}{x}$,we get: $I = \frac{1}{\sqrt{3}} 	an^{-1}(\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$.

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

Similar Questions

$\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1}(x) d x+\int \tan ^{-1}(1-x) d x=$

Find the integral of the function $\tan^{3} 2x \sec 2x$.

If $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$,then $f\left(\frac{2 \pi}{3}\right) =$

$\int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx =$

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$,then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module