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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To evaluate the integral $I = \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$,divide the numerator and denominator by $x^2$: $I = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \,dx}$ Rewrite the denominator as a perfect square: $I = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} ight)}^2} + 2}}} \,dx$ Let $t = x - \frac{1}{x}$. Then $dt = \left( {1 + \frac{1}{{{x^2}}}} ight)dx$. Substituting these into the integral,we get: $I = \int {\frac{{dt}}{{{t^2} + {{(\sqrt 2 )}^2}}}} = \frac{1}{{\sqrt 2 }}{	an ^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} ight) + c$ Substituting $t = x - \frac{1}{x} = \frac{{{x^2} - 1}}{x}$ back into the expression: $I = \frac{1}{{\sqrt 2 }}{	an ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} ight) + c$.

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

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