Mathematics

Evaluate 
$$i)\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1}$$ dx
$$ii)\displaystyle \int \dfrac{dx}{x^2 + 1}$$


SOLUTION
$$i)\displaystyle \int{\cfrac{{x}^{2} + 1}{{x}^{4} + 1} dx}$$
$$= \displaystyle\int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{x}^{2} + \cfrac{1}{{x}^{2}}} dx}$$
$$= \displaystyle \int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{\left( x - \cfrac{1}{x} \right)}^{2} + 2} dx}$$
Let
$$x - \cfrac{1}{x} = t \Rightarrow \left( 1 + \cfrac{1}{{x}^{2}} \right) dx = dt$$
Therefore, the integral  become-
$$\int{\cfrac{dt}{{t}^{2} + {\left( \sqrt{2} \right)}^{2}}}$$
$$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{t}{\sqrt{2}} \right)} + C$$
Substituting the value of $$t$$, we get
$$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{\left( 1 + \cfrac{1}{{x}^{2}} \right)}{\sqrt{2}} \right)} + C$$
$$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{{x}^{2} + 1}{\sqrt{2} {x}^{2}} \right)} + C$$


$$ii)\displaystyle \int{\cfrac{1}{{x}^{2} + 1} dx}$$
$$\displaystyle \int{\cfrac{1}{{x}^{2} + {\left( 1 \right)}^{2}} dx}$$
$$= \tan^{-1}{x} + C$$
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