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Question Number 19666 by Joel577 last updated on 14/Aug/17

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

Answered by ajfour last updated on 14/Aug/17

$$\mathrm{I}=\frac{1}{2}\int {\mathrm{x}}^3\left(2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}\right)\mathrm{dx}$$$$=\frac{{2\mathrm{x}}^3}{3}{(1+{\mathrm{x}}^2)}^{3/2}-\frac{{\mathrm{I}}_1}{2}+{\mathrm{C}}_0....\left(\mathrm{i}\right)$$$${\mathrm{I}}_1=\int \left({3\mathrm{x}}^2\right)\left(\frac{2}{3}\right){(1+{\mathrm{x}}^2)}^{3/2}\mathrm{dx}$$$$\mathrm{let}1+{\mathrm{x}}^2={\mathrm{t}}^2\Rightarrow \mathrm{xdx}=\mathrm{dt}$$$$\mathrm{dx}=\frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2-1}}$$$${\mathrm{I}}_1=2\int ({\mathrm{t}}^2-1)\left({\mathrm{t}}^3\right)\frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2-1}}$$$$=2\int {\mathrm{t}}^3\sqrt{{\mathrm{t}}^2-1}\mathrm{dt}$$$$=\int {\mathrm{t}}^2\left(2\mathrm{t}\sqrt{{\mathrm{t}}^2-1}\right)\mathrm{dt}$$$$=\frac{{2\mathrm{t}}^2}{3}{({\mathrm{t}}^2-1)}^{3/2}-\frac{2}{3}\int \left(2\mathrm{t}\right){({\mathrm{t}}^2-1)}^{3/2}+{\mathrm{C}}_1$$$$=\frac{{2\mathrm{t}}^2}{3}{({\mathrm{t}}^2-1)}^{3/2}-\frac{2}{3}\times \frac{2}{5}{({\mathrm{t}}^2-1)}^{5/2}+{\mathrm{C}}_1+{\mathrm{C}}_2$$$$=\frac{2}{3}({\mathrm{x}}^2+1){\mathrm{x}}^3-\frac{4}{15}{\mathrm{x}}^5+{\mathrm{C}}_1+{\mathrm{C}}_2..\left(\mathrm{ii}\right)$$$$\mathrm{so}\mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right):$$$$\mathrm{I}=\frac{{2\mathrm{x}}^3}{3}{(1+{\mathrm{x}}^2)}^{3/2}-\frac{{\mathrm{x}}^3}{3}(1+{\mathrm{x}}^2)+\frac{{2\mathrm{x}}^5}{15}+\mathrm{C}.$$

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