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

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

Question

(C) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} - 1 + \frac{1}{{{x^2}}}}}\,dx} $ $= \int {\frac{{1 + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} - 1 + \frac{1}{{{x^2}}}}}\,dx} $ $= \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} ight)}^2} + 1}}\,dx} $ Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{{{x^2}}})\,dx$. Substituting these into the integral: $\int {\frac{{dt}}{{{t^2} + 1}} = {{	an }^{ - 1}}t + c} $ Substituting back $t = x - \frac{1}{x}$: $= {	an ^{ - 1}}\left( {x - \frac{1}{x}} ight) + c = {	an ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} ight) + c$.

$\int {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\,dx = }$

Similar Questions

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

$\int {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} \,dx$ equals

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

$\int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx =$

If $\int \frac{dx}{x \sqrt{1-x^3}} = k \log \left(\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right) + c$,(where $c$ is a constant of integration),then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module