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Question Number 96175 by Fikret last updated on 30/May/20

$$\int \frac{dx}{\sqrt{4x^2+4x+3}}=?$$

Commented by bobhans last updated on 30/May/20

$$\int \frac{dx}{2\sqrt{x^2+x+\frac{3}{4}}}=\int \frac{dx}{2\sqrt{{(x+\frac{1}{2})}^2+\frac{1}{2}}}$$$$setx+\frac{1}{2}=\sqrt{\frac{1}{2}}\tan u\Rightarrow \tan \mathrm{u}=\frac{2\mathrm{x}+1}{\sqrt{2}}$$$$=\frac{1}{2}\int \frac{\frac{1}{\sqrt{2}}\sec {}^2\mathrm{u}\mathrm{du}}{\frac{1}{\sqrt{2}}\sec \mathrm{u}}=\frac{1}{2}\ln \mid \sec \mathrm{u}+\tan \mathrm{u}\mid +\mathrm{c}$$$$$$

Answered by mathmax by abdo last updated on 30/May/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{{4\mathrm{x}}^2+4\mathrm{x}+3}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{{\left(2\mathrm{x}\right)}^2+2\left(2\mathrm{x}\right)+1+2}}$$$$=\int \frac{\mathrm{dx}}{\sqrt{{(2\mathrm{x}+1)}^2+2}}{=}_{2\mathrm{x}+1=\sqrt{2}\mathrm{u}}\int \frac{\mathrm{du}}{\sqrt{2}\sqrt{2}\sqrt{1+{\mathrm{u}}^2}}=\frac{1}{2}\mathrm{argsh}\left(\mathrm{u}\right)+\mathrm{c}$$$$=\frac{1}{2}\ln (\mathrm{u}+\sqrt{1+{\mathrm{u}}^2})+\mathrm{c}=\frac{1}{2}\ln (\frac{2\mathrm{x}+1}{\sqrt{2}}+\sqrt{1+{\left(\frac{2\mathrm{x}+1}{\sqrt{2}}\right)}^2})+\mathrm{c}$$

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