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Question Number 96898 by bobhans last updated on 05/Jun/20

Commented by PRITHWISH SEN 2 last updated on 05/Jun/20

$$\frac{1}{6}\int \{\sqrt{{\mathrm{x}}^2+6\mathrm{x}}+\mathrm{x}\}\mathrm{dx}=\frac{1}{6}\int \sqrt{{(\mathrm{x}+3)}^2-9}\mathrm{dx}+\frac{1}{6}\int \mathrm{xdx}$$$$=\frac{(\mathrm{x}+3)}{12}\sqrt{{\mathrm{x}}^2+6\mathrm{x}}-\frac{3}{4}\ln \mid (\mathrm{x}+3)+\sqrt{{\mathrm{x}}^2+6\mathrm{x}}\mid +\frac{{\mathrm{x}}^2}{12}+\mathrm{C}$$

Answered by john santu last updated on 05/Jun/20

$$\mathrm{I}=\int \frac{\sqrt{\mathrm{x}}\mathrm{dx}}{\sqrt{\mathrm{x}}(\sqrt{1+\frac{6}{\mathrm{x}}}-1)}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{1+\frac{6}{\mathrm{x}}}-1}$$$$\mathrm{set}{\mathrm{z}}^2=1+\frac{6}{\mathrm{x}}\mathrm{then}\mathrm{x}=\frac{6}{{\mathrm{z}}^2-1}$$$$\mathrm{dx}=\frac{-12\mathrm{z}}{{({\mathrm{z}}^2-1)}^2}\mathrm{dz}$$$$\mathrm{I}=\int \frac{1}{1-\mathrm{z}}.\frac{(-12\mathrm{z}}{{({\mathrm{z}}^2-1)}^2}\mathrm{dz}$$$$\mathrm{I}=\int \frac{12\mathrm{z}\mathrm{dz}}{(\mathrm{z}-1){({\mathrm{z}}^2-1)}^2}=\int \frac{12\mathrm{z}\mathrm{dz}}{{(\mathrm{z}-1)}^2(\mathrm{z}+1)}$$$$\mathrm{I}=\int \frac{3\mathrm{dz}}{2(\mathrm{z}-1)}+\int \frac{6\mathrm{dz}}{{(\mathrm{z}-1)}^2}-\int \frac{9\mathrm{dz}}{2(\mathrm{z}+1)}$$$$\mathrm{I}=\frac{3}{2}\ln \mid \mathrm{z}-1\mid -\frac{6}{\mathrm{z}-1}-\frac{9}{2}\ln \mid \mathrm{z}+1\mid +\mathrm{c}$$$$\mathrm{I}=\frac{3}{2}\ln \mid \sqrt{\frac{\mathrm{x}+6}{\mathrm{x}}}-1\mid -\frac{6\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+6}}-\frac{9}{2}\ln \mid \sqrt{\frac{\mathrm{x}+6}{\mathrm{x}}}+1\mid +\mathrm{c}$$

Answered by Sourav mridha last updated on 05/Jun/20

$$=\frac{1}{6}[\int \sqrt{{\mathit{x}}^2+6\mathit{x}}+\frac{1}{2}{\mathit{x}}^2]$$$$=\frac{1}{6}[\int \sqrt{{(\mathit{x}+3)}^2-3^2}\mathit{d}(\mathit{x}+3)]+\frac{1}{12}{\mathit{x}}^2$$$$=\frac{1}{6}[\frac{(\mathit{x}+3)}{2}.\sqrt{{\mathit{x}}^2+6\mathit{x}}$$$$-\frac{9}{2}\mathit{l}\mathit{n}[(\mathit{x}+3)+\sqrt{{\mathit{x}}^2+6\mathit{x}}]$$$$+\frac{1}{12}{\mathit{x}}^2+\mathit{c}$$

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