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Question Number 604 by 123456 last updated on 09/Feb/15

$$\underset{0}{\stackrel{+1}{\int }}\left(\underset{-1}{\stackrel{+1}{\int }}\frac{1-x^4{y}^3}{1-x^{2}y}dx\right)dy$$

Answered by prakash jain last updated on 15/Feb/15

$$\frac{1-x^4{y}^3}{1-x^{2}y}=\frac{x^2{y}^2-x^4{y}^3-x^2{y}^2+y+1-y}{1-x^{2}y}$$$$=\frac{x^2{y}^2(1-x^{2}y)+y(1-x^{2}y)+(1-y)}{1-x^{2}y}$$$$=x^2{y}^2+y+\frac{1-y}{1-x^{2}y}$$$$\mathrm{Integrate}\mathrm{wrt}x$$$$\frac{y^2{x}^3}{3}+xy+\frac{1-y}{\sqrt{y}}{\tanh }^{-1}\left(x\sqrt{y}\right)$$$$\mathrm{putting}\mathrm{limits}$$$$=\frac{2y^2}{3}+2y+2\frac{1-y}{\sqrt{y}}{\tanh }^{-1}\left(\sqrt{\mathrm{y}}\right)$$$$\mathrm{I}1$$$${\int }_0^1\frac{2y^2}{3}dy={\left[\frac{2y^3}{9}\right]}_0^1=\frac{2}{9}$$$$\mathrm{I}2$$$${\int }_0^{1}2ydy={\left[y^2\right]}_0^1=1$$$$\mathrm{I}3$$$${\int }_0^{1}2\frac{1-y}{\sqrt{y}}{\tanh }^{-1}\left(\sqrt{\mathrm{y}}\right)dy$$$$y=u^2\Rightarrow dy=2udu$$$${\int }_0^{1}2\frac{1-u^2}{u}{\tanh }^{-1}u2udu$$$$=4{\int }_0^1(1-u^2){\tanh }^{-1}udu$$$$=4[\ln 2-\frac{1}{3}\ln 2-\frac{1}{6}]$$$$=\frac{8}{3}\ln 2-\frac{2}{3}$$$$\mathrm{Final}\mathrm{Answer}$$$$\frac{2}{9}+1+\frac{8}{3}\ln 2-\frac{2}{3}$$$$=\frac{8}{3}\ln 2+\frac{2+9-6}{9}=\frac{8}{3}\ln 2+\frac{5}{9}$$

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