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Question Number 121066 by Jamshidbek2311 last updated on 05/Nov/20

$$1)\int \frac{x}{x+1}dx=?$$$$2)\int \frac{x}{\sqrt{x-1}}dx=?$$$$3)\int x\sqrt{x-4}dx=\stackrel{}{?}$$$$$$

Commented by liberty last updated on 05/Nov/20

$$1)\int \frac{\mathrm{x}+1-1}{\mathrm{x}+1}\mathrm{dx}=\int 1-\frac{1}{\mathrm{x}+1}\mathrm{dx}$$$$=\mathrm{x}-\ln \mid \mathrm{x}+1\mid +\mathrm{c}=\ln {\mathrm{e}}^{\mathrm{x}}-\ln \mid \mathrm{x}+1\mid +\mathrm{c}$$$$=\ln \mid \frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}+1}\mid +\mathrm{c}$$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

$$\int \frac{x}{\sqrt{x-1}}dx$$$$=\int \frac{2tdt}{t}(t^2+1)(x-1)=t$$$$=\frac{2t^3}{3}+2t=\frac{2}{3}{(x-1)}^{\frac{3}{2}}+2\sqrt{x-1}+C$$

Answered by bramlexs22 last updated on 05/Nov/20

$$3)\int \mathrm{x}\sqrt{\mathrm{x}-4}\mathrm{dx}$$$$\mathrm{set}\sqrt{\mathrm{x}-4}=\mathrm{z}\to \mathrm{x}={\mathrm{z}}^2+4$$$$\int ({\mathrm{z}}^2+4)\mathrm{z}\left(2\mathrm{z}\mathrm{dz}\right)=$$$$\int ({\mathrm{z}}^2+4)\left({2\mathrm{z}}^2\right)\mathrm{dz}=$$$$\int {2\mathrm{z}}^4+{8\mathrm{z}}^2\mathrm{dz}=\frac{2}{5}{\mathrm{z}}^5+\frac{8}{3}{\mathrm{z}}^3+\mathrm{c}$$$$={\mathrm{z}}^3(\frac{2}{5}{\mathrm{z}}^2+\frac{8}{3})+\mathrm{c}$$$$={(\mathrm{x}-4)}^{\frac{3}{2}}(\frac{2}{5}(\mathrm{x}-4)+\frac{8}{3})+\mathrm{c}$$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

$$\int x\sqrt{x-4}dx$$$$=\int 2t(t^2+4)tdt{t}^2=x-4$$$$=\frac{2}{5}{t}^5+\frac{8}{3}{t}^3=\frac{2}{5}{(x-4)}^{\frac{5}{2}}+\frac{8}{3}{(x-4)}^{\frac{3}{2}}$$

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