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Question Number 169230 by Giantyusuf last updated on 26/Apr/22

Answered by Mathspace last updated on 26/Apr/22

$$I{=}_{x=3^{\frac{1}{4}}t}\int \frac{\sqrt{3}{t}^2}{3(1+t^4)}{3}^{\frac{1}{4}}dt$$$$=3^{-\frac{1}{2}+\frac{1}{4}}\int \frac{t^2}{1+t^4}dt$$$$=\frac{1}{\left({}^4\sqrt{3}\right)}\int \frac{t^2}{1+t^4}dtwehave$$$$\int \frac{t^2}{1+t^4}dt=\int \frac{1}{t^2+\frac{1}{t^2}}dt$$$$=\frac{1}{2}\int \frac{1-\frac{1}{t^2}+1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$$$$=\frac{1}{2}\int \frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2-2}dt(t+\frac{1}{t}=u)$$$$+\frac{1}{2}\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}(t-\frac{1}{t}=v)$$$$=\frac{1}{2}\int \frac{du}{u^2-2}+\frac{1}{2}\int \frac{dv}{v^2+2}$$$$wehave$$$$\int \frac{du}{u^2-2}=\int \frac{du}{(u-\sqrt{2})(u+\sqrt{2})}$$$$=\frac{1}{2\sqrt{2}}\int (\frac{1}{u-\sqrt{2}}-\frac{1}{u+\sqrt{2}})$$$$=\frac{1}{2\sqrt{2}}ln\mid \frac{u-\sqrt{2}}{u+\sqrt{2}}\mid +c_1$$$$=\frac{1}{2\sqrt{2}}ln\mid \frac{t+\frac{1}{t}-\sqrt{2}\mid }{t+\frac{1}{t}+\sqrt{2}}\mid +c_1$$$$=\frac{1}{2\sqrt{2}}ln\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t-1}\right)+c_1$$$$\int \frac{dv}{v^2+2}{=}_{v=\sqrt{2}z}\int \frac{\sqrt{2}dz}{2(1+z^2)}$$$$=\frac{1}{\sqrt{2}}arctanz+c_2$$$$=\frac{1}{\sqrt{2}}arvtan\left(\frac{v}{\sqrt{2}}\right)+c_2$$$$=\frac{1}{\sqrt{2}}arctan\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c_2$$$$=\frac{1}{\sqrt{2}}artan\left(\frac{x^2-1}{x\sqrt{2}}\right)+c_2\Rightarrow$$

Commented by Mathspace last updated on 26/Apr/22

$$\Rightarrow I=\frac{1}{4\sqrt{2}}ln\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)$$$$+\frac{1}{2\sqrt{2}}arctan\left(\frac{t^2-1}{t\sqrt{2}}\right)+C$$$$t=\frac{x}{\left({}^4\sqrt{3}\right)}$$

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