Question

evaluate integrate integrals

Answer 1

I tried to explain

hope you'll understand

Answer 2

Question :

$\dagger\ \; \displaystyle \sf \red{Evaluate\ \int \dfrac{(x^2+1)}{(x^4+1)}\ dx}$

Solution :

$\displaystyle \sf \int \dfrac{(x^2+1)}{(x^4+1)}\ dx$

Taking 'x²' common in both num. and den. ,

$\longrightarrow \displaystyle \sf \int \dfrac{x^2 \left(1+\frac{1}{x^2}\right)}{x^2 \left( x^2+\frac{1}{x^2} \right) }\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{ \left(1+\frac{1}{x^2}\right)}{ \left( x^2+\frac{1}{x^2} \right) }\ dx$

$\bullet\ \; \sf \orange{(A-B)^2+2AB = A^2+B^2}$

$\longrightarrow \displaystyle \sf \int \dfrac{ \left(1+\frac{1}{x^2}\right)}{ \left( x - \frac{1}{x} \right)^2 + 2.x.\frac{1}{x} }\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{ \left(1+\frac{1}{x^2}\right)}{ \left( x-\frac{1}{x} \right)^2 + 2}\ dx$

Use sub. method ,

$\implies \sf \blue{u= x-\dfrac{1}{x}}$

$\implies \sf \green{du = \left( 1+\dfrac{1}{x^2} \right) dx}$

$\longrightarrow \displaystyle \sf \int \dfrac{ du}{ u^2 + 2}$

Remember the formula ,

$\bullet\ \; \displaystyle \sf \orange{\int \dfrac{dx}{a^2+x^2}= \dfrac{1}{a}\ tan^{-1}\ \dfrac{x}{a}}$

$\longrightarrow \displaystyle \sf \int \dfrac{ du}{ (\sqrt{2})^2 + u^2}$

$\longrightarrow \displaystyle \sf \dfrac{1}{\sqrt{2}}\ tan^{-1}\ \dfrac{u}{\sqrt{2}}+c$

We have ,

$\bullet\ \; \sf \blue{u=x-\dfrac{1}{x}}$

$\longrightarrow \displaystyle \sf \dfrac{1}{\sqrt{2}}\ tan^{-1} \left( \dfrac{x- \frac{1}{x}}{\sqrt{2}} \right) +c$

$\longrightarrow \displaystyle \sf \pink{\dfrac{1}{\sqrt{2}}\ tan^{-1} \left( \dfrac{x^2- 1}{\sqrt{2}\ x} \right) +c}$