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Question Number 122205 by mathmax by abdo last updated on 14/Nov/20

$$\mathrm{find}\int \frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}}}$$

Commented by liberty last updated on 15/Nov/20

$$\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+\mathrm{x})}^{\frac{3}{2}}}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^3{(1+{\mathrm{x}}^{-1})}^{\frac{3}{2}}}$$$$=\int \frac{{\mathrm{x}}^{-2}\mathrm{dx}}{\mathrm{x}{(1+{\mathrm{x}}^{-1})}^{\frac{3}{2}}};[{\mathrm{u}}^{\frac{2}{3}}=1+\frac{1}{\mathrm{x}}]$$$$\Rightarrow [\frac{2}{3}{\mathrm{u}}^{-\frac{1}{3}}\mathrm{du}=-{\mathrm{x}}^{-2}\mathrm{dx}]$$$$\mathrm{J}=\int ({\mathrm{u}}^{\frac{2}{3}}-1)(-\frac{2}{3}{\mathrm{u}}^{-\frac{1}{3}}\mathrm{du})\left(\frac{1}{\mathrm{u}}\right)$$$$\mathrm{J}=-\frac{2}{3}\int ({\mathrm{u}}^{\frac{2}{3}}-1)\left({\mathrm{u}}^{-\frac{4}{3}}\right)\mathrm{du}$$$$\mathrm{J}=-\frac{2}{3}\int ({\mathrm{u}}^{-\frac{2}{3}}-{\mathrm{u}}^{-\frac{4}{3}})\mathrm{du}$$$$\mathrm{J}=-\frac{2}{3}[{3\mathrm{u}}^{\frac{1}{3}}+{3\mathrm{u}}^{-\frac{1}{3}}]+\mathrm{c}$$$$\mathrm{J}=-2\sqrt{\frac{\mathrm{x}+1}{\mathrm{x}}}-2\sqrt{\frac{\mathrm{x}}{\mathrm{x}+1}}+\mathrm{c}$$$$\mathrm{J}=-2\left(\frac{\mathrm{x}+1+\mathrm{x}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}}}\right)+\mathrm{c}=-\frac{4\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+\mathrm{x}}}+\mathrm{c}$$

Commented by benjo_mathlover last updated on 15/Nov/20

$$letJ\left(x\right)=-\frac{4x+2}{\sqrt{x^2+x}}$$$$\frac{dJ\left(x\right)}{dx}=-\left[\frac{4\sqrt{x^2+x}-(4x+2)\left(\frac{2x+1}{2\sqrt{x^2+x}}\right)}{x^2+x}\right]$$$$=-\left[\frac{4\sqrt{x^2+x}-\frac{{(2x+1)}^2}{\sqrt{x^2+x}}}{x^2+x}\right]$$$$=-\left[\frac{4x^2+4x-4x^2-4x-1}{(x^2+x)\sqrt{x^2+x}}\right]$$$$=\frac{1}{x(x+1)\sqrt{x^2+x}}$$$$correctsirLiberty.$$

Answered by TANMAY PANACEA last updated on 14/Nov/20

$$\int \frac{(x+1)-x}{x(x+1)\sqrt{x^2+x}}dx$$$$\int \frac{dx}{x\sqrt{x^2+x}}-\int \frac{dx}{(x+1)\sqrt{x^2+x}}$$$$a=\frac{1}{x}\to dx=\frac{-1}{a^2}dab=\frac{1}{x+1}\to dx=-\frac{1}{b^2}db$$$$\int \frac{-da}{a^2\times \frac{1}{a}\times \sqrt{\frac{1}{a^2}+\frac{1}{a}}}-\int \frac{-db}{b^2\times \frac{1}{b}\sqrt{{(\frac{1}{b}-1)}^2+(\frac{1}{b}-1)}}$$$$\int \frac{-da}{\sqrt{1+a^2}}-\int \frac{db}{b\times \sqrt{\frac{1}{b^2}-\frac{2}{b}+1+\frac{1}{b}-1}}$$$$\int \frac{-da}{\sqrt{1+a^2}}-\int \frac{db}{\sqrt{1-b}}$$$$\int \frac{-da}{\sqrt{1+a^2}}+\int \frac{d(1-b)}{\sqrt{1-b}}$$$$-ln(a+\sqrt{1+a^2})+\frac{{(1-b)}^{\frac{1}{2}}}{\frac{1}{2}}+C$$$$-ln(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}+2{(1-\frac{1}{x+1})}^{\frac{1}{2}}+c$$

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