Question Number 124980 by mathmax by abdo last updated on 07/Dec/20
$$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}−\mathrm{1}}+\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{2}} } \\ $$
Answered by liberty last updated on 07/Dec/20
![[ (((√(x+1))−2(√(x+1)))/(x−1−4(x+1))) ]^2 =[ (((√(x−1))−2(√(x+1)))/(−3x−5)) ]^2 ∫ (((x−1)+4(x+1)−4(√(x^2 −1)))/((3x+5)^2 )) dx = ∫ ((5x+3)/((3x+5)^2 )) dx −4∫ ((√(x^2 −1))/((3x+5)^2 )) dx = I_1 = ∫ (((5/3)(3x+5)−((16)/3))/((3x+5)^2 )) dx = −(5/(3(3x+5)))−((16)/3)(((3x+5)^(−3) )/((−9))) = −(5/(9x+15)) + ((16)/(27(3x+5)^3 )) I_2 = 4∫ ((√(x^2 −1))/((3x+5)^2 )) dx](https://www.tinkutara.com/question/Q124992.png)
$$\: \\ $$$$\left[\:\:\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{2}\sqrt{{x}+\mathrm{1}}}{{x}−\mathrm{1}−\mathrm{4}\left({x}+\mathrm{1}\right)}\:\right]^{\mathrm{2}} \:=\left[\:\frac{\sqrt{{x}−\mathrm{1}}−\mathrm{2}\sqrt{{x}+\mathrm{1}}}{−\mathrm{3}{x}−\mathrm{5}}\:\right]^{\mathrm{2}} \\ $$$$\int\:\frac{\left({x}−\mathrm{1}\right)+\mathrm{4}\left({x}+\mathrm{1}\right)−\mathrm{4}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{2}} }\:{dx}\:= \\ $$$$\int\:\frac{\mathrm{5}{x}+\mathrm{3}}{\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{2}} }\:{dx}\:−\mathrm{4}\int\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{2}} }\:{dx}\:= \\ $$$${I}_{\mathrm{1}} =\:\int\:\frac{\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{3}{x}+\mathrm{5}\right)−\frac{\mathrm{16}}{\mathrm{3}}}{\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{2}} }\:{dx}\:=\:−\frac{\mathrm{5}}{\mathrm{3}\left(\mathrm{3}{x}+\mathrm{5}\right)}−\frac{\mathrm{16}}{\mathrm{3}}\frac{\left(\mathrm{3}{x}+\mathrm{5}\right)^{−\mathrm{3}} }{\left(−\mathrm{9}\right)} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{9}{x}+\mathrm{15}}\:+\:\frac{\mathrm{16}}{\mathrm{27}\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{3}} }\: \\ $$$${I}_{\mathrm{2}} \:=\:\mathrm{4}\int\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{2}} }\:{dx}\: \\ $$