Question-84708

Question Number 84708 by Power last updated on 15/Mar/20

Answered by TANMAY PANACEA last updated on 16/Mar/20

∫(dx/((x+2)^2 (√(x^2 +2x−5)))) ∫(dx/((x+2)^2 (√((x+1)^2 −6)))) x+2=(1/t)→dx=((−dt)/t^2 ) ∫((−dt)/(t^2 ×(1/t^2 )(√(((1/t)−1)^2 −6)))) ∫((−dt)/( (√((1/t^2 )−(2/t)−5)))) ∫((−t dt)/( (√(1−2t−5t^2 )))) (1/(10))∫((−10t−2+2)/( (√(−5t^2 −2t+1))))dt now... −5t^2 −2t+1 5{(((√6)/5))^2 −(t+(1/5))^2 } so (1/(10))∫((d(−5t^2 −2t+1))/( (√(−5t^2 −2t+1))))+(1/5)∫(dt/( (√5) (√((((√6)/5))^2 −(t+(1/5))^2 )) )) (1/(10))×(((−5t^2 −2t+1)^(1/2) )/(1/2))+(1/(5(√5)))sin^(−1) (((t+(1/5))/((√6)/5)))+C pls replace t by (1/(x+2))

$$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}} \\ $$$$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}}} \\ $$$${x}+\mathrm{2}=\frac{\mathrm{1}}{{t}}\rightarrow{dx}=\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\sqrt{\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}}−\mathrm{5}}} \\ $$$$\int\frac{−{t}\:{dt}}{\:\sqrt{\mathrm{1}−\mathrm{2}{t}−\mathrm{5}{t}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}\int\frac{−\mathrm{10}{t}−\mathrm{2}+\mathrm{2}}{\:\sqrt{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}}{dt} \\ $$$${now}… \\ $$$$−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1} \\ $$$$\mathrm{5}\left\{\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}\right)^{\mathrm{2}} −\left({t}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} \right\} \\ $$$$\boldsymbol{{so}} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}\int\frac{{d}\left(−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)}{\:\sqrt{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dt}}{\:\sqrt{\mathrm{5}}\:\sqrt{\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}\right)^{\mathrm{2}} −\left({t}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} }\:} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}×\frac{\left(−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{5}}}{sin}^{−\mathrm{1}} \left(\frac{{t}+\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}}\right)+\boldsymbol{{C}} \\ $$$${pls}\:{replace}\:{t}\:{by}\:\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$ \\ $$

Commented by Power last updated on 15/Mar/20

$$\mathrm{thanks} \\ $$

Commented by TANMAY PANACEA last updated on 15/Mar/20

$${most}\:{welcome} \\ $$

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