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Question Number 35429 by abdo.msup.com last updated on 18/May/18

$$find{\int }_0^1\frac{2x-1}{\sqrt{x^2+6}}dx$$

Commented by prof Abdo imad last updated on 19/May/18

$$letputI={\int }_0^1\frac{2x-1}{\sqrt{x^2+6}}.changementx=\sqrt{6}sht$$$$giveI={\int }_0^{argsh\left(\frac{1}{\sqrt{6}}\right)}\frac{2\sqrt{6}sht-1}{\sqrt{6}ch\left(t\right)}\sqrt{6}chtdt$$$$={\int }_0^{ln(\frac{1}{\sqrt{6}}+\sqrt{1+\frac{1}{6}}}(2\sqrt{6}sh\left(t\right)-1)dt$$$$=2\sqrt{6}{\left[ch\left(t\right)\right]}_0^{ln(\frac{1}{\sqrt{6}}+\sqrt{\frac{7}{6}})}-ln(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})$$$$wehavech\left(t\right)=\frac{e^t+e^{-t}}{2}\Rightarrow$$$$ch\left\{ln(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})\right\}=\frac{(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})+{(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})}^{-1}}{2}\Rightarrow$$$$I=2\sqrt{6}\{\frac{(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})+{(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})}^{-1}}{2}-1\}$$$$-ln(\frac{1}{\sqrt{6}}+\frac{\sqrt{7}}{\sqrt{6}})$$

Answered by ajfour last updated on 19/May/18

$$I={\int }_0^1\frac{2xdx}{\sqrt{x^2+6}}-{\int }_0^1\frac{dx}{\sqrt{x^2+{\left(\sqrt{6}\right)}^2}}$$$$=2\sqrt{x^2+6}{\mid }_0^1-\ln \mid x+\sqrt{x^2+6}{\mid }_0^1$$$$\Rightarrow I=2(\sqrt{7}-\sqrt{6})-\ln \left(\frac{1+\sqrt{7}}{\sqrt{6}}\right)$$

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