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Question Number 102128 by Study last updated on 06/Jul/20

$$\int \frac{x+1}{\sqrt{x}+1}dx=?$$

Answered by Dwaipayan Shikari last updated on 06/Jul/20

$$\int \frac{x}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}=\int \frac{2u^3}{u+1}du+\int \frac{2u}{u+1}du\{takexas{u}^2$$$$=2\int \frac{u^3+1}{u+1}-\frac{1}{u+1}du+2\int 1-\frac{1}{u+1}du$$$$=2\int u^2-u+1-2log(u+1)+2u-2log(u+1)$$$$=2\int {(u-1)}^2+udu+2u-4log(u+1)$$$$=2\int {(u-1)}^{2}du+u^2+2u-4log(u+1)$$$$=\frac{2}{3}{(u-1)}^3+u^2+2u-4log(u+1)+C$$$$=\frac{2}{3}{(\sqrt{x}-1)}^3+x+2\sqrt{x}-4log(\sqrt{x}+1)+C$$$$$$

Answered by abdomathmax last updated on 06/Jul/20

$$\mathrm{I}=\int \frac{\mathrm{x}+1}{\sqrt{\mathrm{x}}+1}\mathrm{dx}\mathrm{changement}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{t}}^2\Rightarrow$$$$\mathrm{I}=\int \frac{{\mathrm{t}}^2+1}{\mathrm{t}+1}\left(2\mathrm{t}\right)\mathrm{dt}=2\int \frac{{\mathrm{t}}^3+\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=2\int \frac{(\mathrm{t}+1-1)\left({\mathrm{t}}^2\right)+\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=2\int {\mathrm{t}}^2\mathrm{dt}+2\int \frac{-{\mathrm{t}}^2+\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=\frac{2}{3}{\mathrm{t}}^3-2\int \frac{{\mathrm{t}}^2-\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}\mathrm{but}$$$$\int \frac{{\mathrm{t}}^2-\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}=\int \frac{\mathrm{t}(\mathrm{t}+1-1)-\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=\int \mathrm{t}\mathrm{dt}-2\int \frac{\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=\frac{{\mathrm{t}}^2}{2}-2\int \frac{\mathrm{t}+1-1}{\mathrm{t}+1}\mathrm{dt}=\frac{{\mathrm{t}}^2}{2}-2\mathrm{t}+2\ln \mid \mathrm{t}+1\mid +\mathrm{c}\Rightarrow$$$$\mathrm{I}=\frac{2}{3}{\mathrm{t}}^3-{\mathrm{t}}^2+4\mathrm{t}-4\ln \mid \mathrm{t}+1\mid +\mathrm{C}$$$$\mathrm{I}=\frac{2}{3}{\left(\sqrt{\mathrm{x}}\right)}^3-\mathrm{x}+4\sqrt{\mathrm{x}}-4\ln \mid \sqrt{\mathrm{x}}+1\mid +\mathrm{C}$$

Commented by mathmax by abdo last updated on 06/Jul/20

$$\mathrm{do}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{get}\mathrm{another}\mathrm{answer}...!$$

Commented by Dwaipayan Shikari last updated on 07/Jul/20

$$Sorrysir!$$

Answered by Ar Brandon last updated on 06/Jul/20

$$\mathcal{I}=\int \frac{\mathrm{x}+1}{\sqrt{\mathrm{x}}+1}\mathrm{dx},\mathrm{Let}\mathrm{t}=\sqrt{\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{x}+1}{\sqrt{\mathrm{x}}+1}=\frac{{\mathrm{t}}^2+1}{\mathrm{t}+1}=\mathrm{t}-1+\frac{2}{\mathrm{t}+1}$$$$\Rightarrow \mathcal{I}=\int \{\sqrt{\mathrm{x}}-1+\frac{2}{\sqrt{\mathrm{x}}+1}\}\mathrm{dx}=\int \{{\mathrm{x}}^{\frac{1}{2}}-1+\frac{2(\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}\}\mathrm{dx}$$$$=\int \{{\mathrm{x}}^{\frac{1}{2}}-1-\frac{2}{\mathrm{x}-1}\}\mathrm{dx}+2\int \frac{\sqrt{\mathrm{x}}}{\mathrm{x}-1}\mathrm{dx},{\mathrm{u}}^2=\mathrm{x}\Rightarrow 2\mathrm{udu}=\mathrm{dx}$$$$=\frac{{2\mathrm{x}}^{\frac{3}{2}}}{3}-\mathrm{x}-2\ln \mid \mathrm{x}-1\mid +2\int \frac{\mathrm{u}}{{\mathrm{u}}^2-1}\cdot 2\mathrm{udu}$$$$=\frac{{2\mathrm{x}}^{\frac{3}{2}}}{3}-\mathrm{x}-2\ln \mid \mathrm{x}-1\mid +4\int \frac{{\mathrm{u}}^2}{{\mathrm{u}}^2-1}\mathrm{du}$$$$\mathcal{K}=\int \frac{{\mathrm{u}}^2}{{\mathrm{u}}^2-1}\mathrm{du}=\int \{1+\frac{1}{{\mathrm{u}}^2-1}\}\mathrm{du}=\mathrm{u}-{\tanh }^{-1}\left(\mathrm{u}\right)$$$$\Rightarrow \mathcal{I}=\frac{{2\mathrm{x}}^{\frac{3}{2}}}{3}-\mathrm{x}-2\ln \mid \mathrm{x}-1\mid +\sqrt{\mathrm{x}}-{\tanh }^{-1}\left(\sqrt{\mathrm{x}}\right)$$$$=\frac{2\sqrt{{\mathrm{x}}^3}}{\sqrt{3}}+\sqrt{\mathrm{x}}-\mathrm{x}-2\ln \mid \mathrm{x}-1\mid -{\tanh }^{-1}\left(\sqrt{\mathrm{x}}\right)+\mathcal{C}$$

Answered by OlafThorendsen last updated on 07/Jul/20

$$u=\sqrt{x}+1\Rightarrow dx=2(u-1)du$$$$\int \frac{{(u-1)}^2+1}{u}2(u-1)du$$$$\int 2\frac{u^2-2u+2}{u}(u-1)du$$$$\int 2(u^2-2u+2)(1-\frac{1}{u})du$$$$\int 2(u^2-2u+2-u+2-\frac{2}{u})du$$$$\int 2(u^2-3u+4-\frac{2}{u})du$$$$2(\frac{u^3}{3}-\frac{3}{2}{u}^2+4u-2\ln \mid u\mid )+\mathrm{C}$$$$\frac{2}{3}{(\sqrt{x}+1)}^3-3{(\sqrt{x}+1)}^2+8(\sqrt{x}+1)-4\ln (\sqrt{x}+1)+\mathrm{C}$$

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