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Question Number 13360 by tawa tawa last updated on 19/May/17

∫  (x/(√(3 − 2x − x^2 ))) dx

$$\int\:\:\frac{\mathrm{x}}{\sqrt{\mathrm{3}\:−\:\mathrm{2x}\:−\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$

Answered by ajfour last updated on 19/May/17

I=∫((x+1−1)/(√(4−(x+1)^2 ))) dx  −(1/2)∫((−2(x+1))/(√(4−(x+1)^2 ))) dx−∫(dx/(√(4−(x+1)^2 )))  for first integral let t=4−(x+1)^2   dt=−2(x+1)dx  I=−(1/2)∫(dt/(√t))−sin^(−1) (((x+1)/2))+C_1   I=−(1/2)(2(√t))−sin^(−1) (((x+1)/2))+C  I=−(√(4−(x+1)^2 ))−sin^(−1) (((x+1)/2))+C .

$$\boldsymbol{{I}}=\int\frac{\mathrm{x}+\mathrm{1}−\mathrm{1}}{\sqrt{\mathrm{4}−\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}\:\mathrm{dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{2}\left({x}+\mathrm{1}\right)}{\sqrt{\mathrm{4}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:{dx}−\int\frac{{dx}}{\sqrt{\mathrm{4}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${for}\:{first}\:{integral}\:{let}\:\boldsymbol{{t}}=\mathrm{4}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${dt}=−\mathrm{2}\left({x}+\mathrm{1}\right){dx} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\sqrt{{t}}}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+{C}_{\mathrm{1}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\sqrt{{t}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+{C} \\ $$$${I}=−\sqrt{\mathrm{4}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+{C}\:. \\ $$

Commented by tawa tawa last updated on 19/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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