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Question Number 34225 by abdo imad last updated on 03/May/18

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

Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18

$$\int \frac{1}{x^4+x^2+1}dx$$$$\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}$$$$I.=\frac{1}{2}\int \frac{(1+\frac{1}{x^2})-(1-\frac{1}{x^2})}{x^2+\frac{1}{x^2}+1}dx$$$$I_{1=}\frac{1}{2}\int \frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+3}dx$$$$=\frac{1}{2}\int \frac{{dt}_1}{t_1^2+3}useformula$$$$=\frac{1}{2}.\frac{1}{\sqrt{3}}.{tan}^{-1}\left(\frac{t_1}{\sqrt{3}}\right)$$$$I_2=\frac{1}{2}\int \frac{(1-\frac{1}{x^2})}{{(x+\frac{1}{x})}^2-1}$$$$I_2=\frac{1}{2}.\int \frac{{dt}_2}{t_2^2-1}useformula$$$$I_2=\frac{1}{2}.\frac{1}{2}.ln\mid \frac{1-t_2}{1+t_2}\mid$$$$I=I_1-I_2$$

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