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Question Number 49829 by mhozhez last updated on 11/Dec/18

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

Commented by maxmathsup by imad last updated on 11/Dec/18

$$letI=\int \frac{x^2}{x^4+1}dxvhangementx=\frac{1}{t}give$$$$I=\int \frac{1}{t^2(\frac{1}{t^4}+1)}-\frac{dt}{t^2}=-\int \frac{dt}{t^4(1+\frac{1}{t^4})}=\int \frac{dt}{t^4+1}letdecompose$$$$F\left(t\right)=\frac{1}{1+t^4}\Rightarrow F\left(t\right)=\frac{1}{{(t^2+1)}^2-2t^2}=\frac{1}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}$$$$=\frac{at+b}{t^2+\sqrt{2}t+1}+\frac{ct+d}{t^2-\sqrt{2}t+1}wehaveF(-t)=F\left(t\right)\Rightarrow$$$$\frac{-at+b}{t^2-\sqrt{2}t+1}+\frac{-ct+d}{t^2+\sqrt{2}t+1}=F\left(t\right)\Rightarrow c=-aandb=d\Rightarrow$$$$F\left(t\right)=\frac{at+b}{t^2+\sqrt{2}t+1}+\frac{-at+b}{t^2-\sqrt{2}t+1}$$$$F\left(0\right)=1=2b\Rightarrow b=\frac{1}{2}$$$$F\left(1\right)=\frac{1}{2}=\frac{a+b}{2+\sqrt{2}}+\frac{-a+b}{2-\sqrt{2}}\Rightarrow \frac{(2-\sqrt{2})(a+b)+(2+\sqrt{2})(-a+b)}{2}=\frac{1}{2}\Rightarrow$$$$(2-\sqrt{2}-2-\sqrt{2})a+(2-\sqrt{2}+2+\sqrt{2})b=1\Rightarrow$$$$-2\sqrt{2}a+4.\frac{1}{2}=1\Rightarrow -2\sqrt{2}a=-1\Rightarrow a=\frac{1}{2\sqrt{2}}\Rightarrow$$$$F\left(x\right)=\frac{\frac{1}{2\sqrt{2}}t+\frac{1}{2}}{t^2+\sqrt{2}t+1}+\frac{-\frac{1}{2\sqrt{2}}t+\frac{1}{2}}{t^2-\sqrt{2}t+1}\Rightarrow$$$$\int F\left(x\right)dx=\frac{1}{2\sqrt{2}}\int \frac{t+\sqrt{2}}{t^2+\sqrt{2}t+1}dt-\frac{1}{2\sqrt{2}}\int \frac{t-\sqrt{2}}{t^2-\sqrt{2}t+1}dt=H-K$$$$2\sqrt{2}H=\frac{1}{2}\int \frac{2t+\sqrt{2}+\sqrt{2}}{t^2+\sqrt{2}t+1}dt=\frac{1}{2}ln(t^2+\sqrt{2}t+1)+\frac{\sqrt{2}}{2}\int \frac{dt}{t^2+\sqrt{2}t+1}$$$$\int \frac{dt}{t^2+\sqrt{2}t+1}=\int \frac{dt}{t^2+2\frac{\sqrt{2}}{2}t+\frac{1}{2}+1-\frac{1}{2}}=\int \frac{dt}{{(t+\frac{\sqrt{2}}{2})}^2+\frac{1}{2}}$$$${=}_{t+\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}u}2\int \frac{1}{1+u^2}\frac{du}{\sqrt{2}}=\sqrt{2}arctan\left(\frac{t\sqrt{2}+1}{\sqrt{2}}\right)+c\Rightarrow$$$$2\sqrt{2}H=\frac{1}{2}ln(t^2+\sqrt{2}t+1)+arctan\left(\frac{t\sqrt{2}+1}{\sqrt{2}}\right)+cbutt=\frac{1}{x}\Rightarrow$$$$H=\frac{1}{4\sqrt{2}}ln(\frac{1}{x^2}+\frac{\sqrt{2}}{x}+1)+\frac{1}{2\sqrt{2}}arctan(\frac{1}{x}+\frac{1}{\sqrt{2}})+cwefollowthesame$$$$mannerforcalculusofK.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

$$\int \frac{dx}{x^2+\frac{1}{x^2}}$$$$=\frac{1}{2}\int \frac{1-\frac{1}{x^2}+1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2})}dx$$$$=\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}+\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}$$$$=\frac{1}{2}\times \frac{1}{2\sqrt{2}}ln\left\{\frac{(x+\frac{1}{x})-\sqrt{2}}{(x+\frac{1}{x})+\sqrt{2}}\right\}+\frac{1}{2}\times \frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$$$$plscheck...$$

Commented by mhozhez last updated on 11/Dec/18

$$thankssir$$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

$$mostwelcome...$$

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