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Question Number 86626 by sakeefhasan05@gmail.com last updated on 29/Mar/20

$$\int \sqrt{{\mathrm{x}}^2+4}\mathrm{dx}$$$$\mathrm{answer}\mathrm{quick}\mathrm{pls}$$

Commented by sakeefhasan05@gmail.com last updated on 30/Mar/20

Commented by mathmax by abdo last updated on 30/Mar/20

$$letA=\int \sqrt{x^2+4}dxchangementx=2sh\left(t\right)give$$$$A=\int 2ch\left(t\right)2ch\left(t\right)dt=4\int {ch}^{2}tdt=2\int (1+ch\left(2t\right))dt$$$$=2t+sh\left(2t\right)+C=2t+2sh\left(t\right)ch\left(t\right)+C$$$$=2argsh\left(\frac{x}{2}\right)+x\sqrt{1+\frac{x^2}{4}}+C$$$$=2ln(\frac{x}{2}+\sqrt{1+\frac{x^2}{4}})+\frac{x}{2}\sqrt{x^2+4}+C$$$$=2ln(x+\sqrt{x^2+4})+\frac{1}{2}x\sqrt{x^2+4}+c$$

Answered by TANMAY PANACEA. last updated on 29/Mar/20

$$\sqrt{x^2+4}\int dx-\int [\frac{d\left(\sqrt{x^2+4}\right)}{dx}\int dx]dx$$$$x\sqrt{x^2+4}-\int \frac{2x\times x}{2\sqrt{x^2+4}}dx$$$$x\sqrt{x^2+4}-\int \frac{x^2+4-4}{\sqrt{x^2+4}}dx$$$$x\sqrt{x^2+4}-\int \sqrt{x^2+4}+4\int \frac{dx}{\sqrt{x^2+4}}$$$$2I=x\sqrt{x^2+4}+4ln(x+\sqrt{x^2+4)}$$$$I=\frac{x\sqrt{x^2+4}}{2}+2ln(x+\sqrt{x^2+4})$$

Commented by sakeefhasan05@gmail.com last updated on 29/Mar/20

$$\mathrm{thank}\mathrm{u}\mathrm{very}\mathrm{much}$$$$\mathrm{can}\mathrm{u}\mathrm{pls}\mathrm{do}\mathrm{it}\mathrm{again}\mathrm{with}\mathrm{another}$$$$\mathrm{method}(\mathrm{x}=2\tan \theta )\mathrm{substition}$$

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