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Question Number 66308 by mathmax by abdo last updated on 12/Aug/19

$$find\int \frac{dx}{(x+3)\sqrt{-x^2-4x}}$$

Commented by prof Abdo imad last updated on 16/Aug/19

$$letA=\int \frac{dx}{(x+3)\sqrt{-x^2-4x}}wehave$$$$-x^2-4x=-(x^2+4x+4-4)=4-{(x+2)}^2$$$$weusethechangementx+2=2sint\Rightarrow$$$$A=\int \frac{2costdt}{(2sint-2+3)2cost}=\int \frac{dt}{2sint+1}$$$${=}_{tan\left(\frac{t}{2}\right)=u}\int \frac{1}{(2\frac{2u}{1+u^2}+1)}\frac{2du}{1+u^2}$$$$=\int \frac{2du}{(4u+1+u^2)}=\int \frac{2du}{u^2+4u+1}$$$$u^2+4u+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-1=3\Rightarrow$$$$u_1=-2+\sqrt{3}and{u}_2=-2-\sqrt{3}\Rightarrow$$$$A=\int \frac{2du}{(u-u_1)(u-u_2)}=\frac{1}{\sqrt{3}}\int (\frac{1}{u-u_1}-\frac{1}{u-u_2})du$$$$=\frac{1}{\sqrt{3}}ln\mid \frac{u-u_1}{u-u_2}\mid +C=\frac{1}{\sqrt{3}}ln\mid \frac{tan\left(\frac{t}{2}\right)+2-\sqrt{3}}{tan\left(\frac{t}{2}\right)+2+\sqrt{3}}\mid +C$$$$butt=arcsin\left(\frac{x+2}{2}\right)\Rightarrow$$$$A=\frac{1}{\sqrt{3}}ln\mid \frac{tan(\frac{1}{2}arcsin(\frac{x}{2}+1)+2-\sqrt{3}}{tan\left(\frac{1}{2}arcsin(\frac{x}{2}+1)\right)+2+\sqrt{3}}\mid +C$$$$$$$$$$

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