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Question Number 190812 by Mahliyo last updated on 12/Apr/23

Commented by Frix last updated on 12/Apr/23

$$\mathrm{Use}3\mathrm{steps}$$$$1.t=x+\frac{1}{2}$$$$2.u={\sin }^{-1}\frac{2t}{\sqrt{5}}$$$$3.v=\tan \frac{u}{2}$$

Answered by ARUNG_Brandon_MBU last updated on 12/Apr/23

$$I=\int \frac{xdx}{(1+x)\sqrt{1-x-x^2}}$$$$=\int \frac{dx}{\sqrt{1-x-x^2}}-\int \frac{dx}{(x+1)\sqrt{1-x-x^2}},t=\frac{1}{x+1}$$$$=\int \frac{dx}{\sqrt{\frac{5}{4}-{(x+\frac{1}{2})}^2}}+\int \frac{dt}{t\sqrt{1+\frac{1}{t}-\frac{1}{t^2}}}$$$$=\arcsin \left(\frac{2x+1}{\sqrt{5}}\right)+\int \frac{dt}{\sqrt{t^2+t-1}}$$$$=\arcsin \left(\frac{2x+1}{\sqrt{5}}\right)+\mathrm{argch}\left(\frac{2t+1}{\sqrt{5}}\right)+C$$$$=\arcsin \left(\frac{2x+1}{\sqrt{5}}\right)+\ln \mid \frac{x+3}{\sqrt{5}(x+1)}+\frac{2\sqrt{1-x-x^2}}{\sqrt{5}(x+1)}\mid +C$$

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