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Question Number 120040 by bramlexs22 last updated on 28/Oct/20

$$\int \frac{t^5}{\sqrt{2+t^2}}dt$$

Answered by john santu last updated on 28/Oct/20

$$let2+t^2=h^2\to tdt=hdh$$$$\int \frac{t^{4}tdt}{\sqrt{2+t^2}}=\int \frac{{(h^2-2)}^{2}hdh}{h}$$$$\int (h^4-4h^2+4)dh=\frac{1}{5}{h}^5-\frac{4}{3}{h}^3+4h+c$$$$=\frac{1}{15}\sqrt{2+t^2}\{3{(2+t^2)}^2-20(2+t^2)+60\}+c$$$$=\frac{\sqrt{2+t^2}}{15}\{3t^4-8t^2+32\}+c$$

Commented by Lordose last updated on 28/Oct/20

FASTEST

Commented by bramlexs22 last updated on 29/Oct/20

$$waw...$$

Answered by Lordose last updated on 28/Oct/20

$$\mathrm{set}\mathrm{t}=\sqrt{2}\mathrm{tanx}\Rightarrow \mathrm{dt}=\sqrt{2}(1+{\mathrm{t}}^2)\mathrm{dx}$$$$\int \frac{{\left(\sqrt{2}\right)}^6{\tan }^5{\mathrm{xsec}}^2\mathrm{x}}{\sqrt{2}\mathrm{secx}}\mathrm{dx}$$$$4\sqrt{2}\int {\tan }^5\mathrm{xsecxdx}$$$$4\sqrt{2}\int \mathrm{tanxsecx}{({\sec }^2\mathrm{x}-1)}^2\mathrm{dx}$$$$\mathrm{u}=\mathrm{secx}\Rightarrow \mathrm{du}=\mathrm{secxtanxdx}$$$$4\sqrt{2}\int {({\mathrm{u}}^2-1)}^2\mathrm{du}$$$$4\sqrt{2}\int ({\mathrm{u}}^4-{2\mathrm{u}}^2+1)\mathrm{du}$$$$4\sqrt{2}(\frac{{\mathrm{u}}^5}{5}-\frac{{2\mathrm{u}}^3}{3}+\mathrm{u})+\mathrm{C}$$$$4\sqrt{2}(\frac{{\sec }^5\mathrm{x}}{5}-\frac{{2\sec }^3\left(\mathrm{x}\right)}{3}+\mathrm{secx})+\mathrm{C}$$$$\mathrm{x}={\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)$$$$\mathrm{I}=\frac{1}{15}\sqrt{{\mathrm{t}}^2+2}({3\mathrm{t}}^2-8\mathrm{t}+32)+\mathrm{C}$$$$$$

Commented by bramlexs22 last updated on 29/Oct/20

$$yes$$

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