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Question Number 119644 by bemath last updated on 25/Oct/20

$$\left(1\right)\int x^5\sqrt{x^3+1}dx$$$$\left(2\right)\frac{dy}{dx}={(3x-2y+1)}^2$$

Answered by benjo_mathlover last updated on 26/Oct/20

$$\left(1\right)\int x^3\left(x^2\sqrt{x^3+1}\right)dx$$$$D.Imethod\left(bypart\right)\{\begin{array}{l}u=x^3\to du=3x^2\\ v=\frac{1}{3}\int 3x^2\sqrt{x^3+1}dx=\frac{2}{9}{(x^3+1)}^{\frac{3}{2}}\end{array}$$$$=\frac{2}{9}{x}^3{(x^3+1)}^{\frac{3}{2}}-\frac{2}{9}\int \left(3x^2\right){(x^3+1)}^{\frac{3}{2}}dx$$$$=\frac{2}{9}{x}^3{(x^3+1)}^{\frac{3}{2}}-\frac{4}{45}{(x^3+1)}^{\frac{5}{2}}+C$$$$=\frac{2}{9}{(x^3+1)}^{\frac{3}{2}}(x^3-\frac{2}{5}(x^3+1))+C$$$$=\frac{2}{45}{(x^3+1)}^{\frac{3}{2}}(3x^3-2)+C$$

Answered by 1549442205PVT last updated on 26/Oct/20

$$\mathrm{Put}1+{\mathrm{x}}^3={\mathrm{t}}^2\Rightarrow 2\mathrm{tdt}={3\mathrm{x}}^2\mathrm{dx}$$$$\mathrm{dx}=\frac{2\mathrm{tdt}}{3{\mathrm{x}}^2},{\mathrm{x}}^5\sqrt{{\mathrm{x}}^3+1}\mathrm{dx}=\frac{2}{3}{\mathrm{t}}^2({\mathrm{t}}^2-1)\mathrm{dt}$$$$\mathrm{F}=\frac{2}{3}\int \left({\mathrm{t}}^2({\mathrm{t}}^2-1)\right)\mathrm{dt}=\frac{2}{3}\int ({\mathrm{t}}^4-{\mathrm{t}}^2)\mathrm{dt}$$$$=\frac{{2\mathrm{t}}^5}{15}-\frac{2}{9}{\mathrm{t}}^3$$$$=\frac{2{(1+{\mathrm{x}}^3)}^2\sqrt{1+{\mathrm{x}}^3}}{15}-\frac{2}{9}(1+{\mathrm{x}}^3)\sqrt{1+{\mathrm{x}}^3}+\mathrm{C}$$

Commented by benjo_mathlover last updated on 26/Oct/20

$$itshouldbe\frac{2}{3}(\frac{1}{5}{t}^5-\frac{1}{3}{t}^3)=\frac{2}{15}{t}^5-\frac{2}{9}{t}^3$$

Commented by 1549442205PVT last updated on 26/Oct/20

$$\mathrm{Ok},\mathrm{a}\mathrm{mistake}.\mathrm{I}\mathrm{corrected}.\mathrm{Thank}\mathrm{sir}$$

Answered by benjo_mathlover last updated on 26/Oct/20

$$\left(2\right)set3x-2y+1=zand3-2\frac{dy}{dx}=\frac{dz}{dx}$$$$\Rightarrow \frac{dy}{dx}=\frac{3}{2}-\frac{1}{2}\frac{dz}{dx}$$$$\Rightarrow \frac{3}{2}-\frac{1}{2}\frac{dz}{dx}=z^2;\frac{dz}{dx}=3-2z^2$$$$\int \frac{dz}{3-2z^2}=\int dx;\frac{1}{2\sqrt{3}}\int (\frac{1}{\sqrt{3}-z\sqrt{2}}+\frac{1}{\sqrt{3}+z\sqrt{2}})dz=x+c$$$$\frac{1}{2\sqrt{3}}(-\frac{1}{\sqrt{2}}\ln (\sqrt{3}-2\sqrt{2})+\frac{1}{\sqrt{2}}\ln (\sqrt{3}+z\sqrt{2}))=x+c$$$$\frac{1}{2\sqrt{6}}\ln \left(\frac{\sqrt{3}+z\sqrt{2}}{\sqrt{3}-z\sqrt{2}}\right)=x+c$$$$\frac{1}{2\sqrt{6}}\ln \left(\frac{\sqrt{3}+\sqrt{2}(3x-2y+1)}{\sqrt{3}-\sqrt{2}(3x-2y+1)}\right)=x+c$$$$\ln \left(\frac{\sqrt{3}+\sqrt{2}(3x-2y+1)}{\sqrt{3}-\sqrt{2}(3x-2y+1)}\right)=2x\sqrt{6}+C$$$$\iff \frac{\sqrt{3}+\sqrt{2}(3x-2y+1)}{\sqrt{3}-\sqrt{2}(3x-2y+1)}=\lambda e^{2x\sqrt{6}}$$

Answered by TANMAY PANACEA last updated on 26/Oct/20

$$1)t^2=x^3+1\to \frac{2}{3}tdt=x^{2}dx$$$$\int x^3\sqrt{x^3+1}\left(x^{2}dx\right)$$$$\int (t^2-1)\left(t\right)\left(\frac{2}{3}tdt\right)$$$$\frac{2}{3}\int t^4-t^{2}dt$$$$\frac{2}{3}(\frac{t^5}{5}-\frac{t^3}{3})+c$$$$\frac{2}{15}{(x^3+1)}^{\frac{5}{2}}-\frac{2}{9}{(x^3+1)}^{\frac{3}{2}}+c$$$$$$

Answered by Olaf last updated on 26/Oct/20

$$\left(2\right)$$$$u=3x-2y+1$$$$\frac{du}{dx}=3-2\frac{dy}{dx}$$$$\frac{dy}{dx}=-\frac{1}{2}.\frac{du}{dx}+\frac{3}{2}$$$$$$$$-\frac{1}{2}.\frac{du}{dx}+\frac{3}{2}=u^2$$$$\frac{du}{dx}=3-2u^2$$$$\frac{du}{3-2u^2}=dx$$$$\frac{du}{(\sqrt{3}-\sqrt{2}u)(\sqrt{3}+\sqrt{2}u)}=dx$$$$\frac{1}{2\sqrt{3}}[\frac{1}{\sqrt{3}-\sqrt{2}u}+\frac{1}{\sqrt{3+}\sqrt{2}u}]du=dx$$$$\frac{1}{2\sqrt{3}}[-\frac{1}{\sqrt{2}}\ln \mid \sqrt{3}-\sqrt{2}u\mid +\frac{1}{\sqrt{2}}\ln \mid \sqrt{3}+\sqrt{2}u\mid ]=x+{\mathrm{C}}_1$$$$\ln \mid \frac{\sqrt{3}+\sqrt{2}u}{\sqrt{3}-\sqrt{2}u}\mid =2\sqrt{6}x+{\mathrm{C}}_2$$$$\mid \frac{\sqrt{3}+\sqrt{2}u}{\sqrt{3}-\sqrt{2}u}\mid ={\mathrm{C}}_3{e}^{2\sqrt{6}x}$$$$\frac{\sqrt{3}+\sqrt{2}u}{\sqrt{3}-\sqrt{2}u}={\mathrm{C}}_4{e}^{2\sqrt{6}x}$$$$\sqrt{3}+\sqrt{2}u=(\sqrt{3}-\sqrt{2}u){\mathrm{C}}_4{e}^{2\sqrt{6}x}$$$$\sqrt{2}u(1+{\mathrm{C}}_4{e}^{2\sqrt{6}x})=\sqrt{3}(-1+{\mathrm{C}}_4{e}^{2\sqrt{6}x})$$$$u=\sqrt{\frac{3}{2}}\left(\frac{-1+{\mathrm{C}}_4{e}^{2\sqrt{6}x}}{1+{\mathrm{C}}_4{e}^{2\sqrt{6}x}}\right)=3x-2y+1$$$$y=-\frac{1}{2}\sqrt{\frac{3}{2}}\left(\frac{-1+{\mathrm{C}}_4{e}^{2\sqrt{6}x}}{1+{\mathrm{C}}_4{e}^{2\sqrt{6}x}}\right)+\frac{3}{2}x+\frac{1}{2}$$

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