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Question Number 88307 by M±th+et£s last updated on 09/Apr/20

∫(x^2 /(x^2 −(5/2)x−(3/2))) dx

$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}{x}−\frac{\mathrm{3}}{\mathrm{2}}}\:{dx} \\ $$

Answered by TANMAY PANACEA. last updated on 09/Apr/20

∫((2x^2 )/(2x^2 −5x−3)) ∫((2x^2 )/(2x^2 −6x+x−3))dx 2∫(x^2 /(2x(x−3)+1(x−3)))dx ∫((2x^2 dx)/((x−3)(2x+1))) ((2x^2 )/((x−3)(2x+1)))=1+(a/(x−3))+(b/(2x+1)) 2x^2 =2x^2 −5x−3+2ax+a+bx−b 0=x(2a+b−5)+(a−b−3) 2a+b−5=0 a−b−3=0 3a=8→a=(8/3) b=(8/3)−3=((−1)/3) ∫dx+(8/3)∫(dx/(x−3))+((−1)/3)∫(dx/(2x+1)) ∫dx+(8/3)∫(dx/(x−3))+((−1)/6)∫(dx/(x+(1/2))) x+(8/3)ln(x−3)+((−1)/6)ln(x+(1/2))+c

$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}} \\ $$$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+{x}−\mathrm{3}}{dx} \\ $$$$\mathrm{2}\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}\left({x}−\mathrm{3}\right)+\mathrm{1}\left({x}−\mathrm{3}\right)}{dx} \\ $$$$\int\frac{\mathrm{2}{x}^{\mathrm{2}} {dx}}{\left({x}−\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}−\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}=\mathrm{1}+\frac{{a}}{{x}−\mathrm{3}}+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}+\mathrm{2}{ax}+{a}+{bx}−{b} \\ $$$$\mathrm{0}={x}\left(\mathrm{2}{a}+{b}−\mathrm{5}\right)+\left({a}−{b}−\mathrm{3}\right) \\ $$$$\mathrm{2}{a}+{b}−\mathrm{5}=\mathrm{0} \\ $$$${a}−{b}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}{a}=\mathrm{8}\rightarrow{a}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${b}=\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{3}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\int{dx}+\frac{\mathrm{8}}{\mathrm{3}}\int\frac{{dx}}{{x}−\mathrm{3}}+\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\int{dx}+\frac{\mathrm{8}}{\mathrm{3}}\int\frac{{dx}}{{x}−\mathrm{3}}+\frac{−\mathrm{1}}{\mathrm{6}}\int\frac{{dx}}{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}+\frac{\mathrm{8}}{\mathrm{3}}{ln}\left({x}−\mathrm{3}\right)+\frac{−\mathrm{1}}{\mathrm{6}}{ln}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{c} \\ $$

Commented by M±th+et£s last updated on 09/Apr/20

god bless you

$${god}\:{bless}\:{you} \\ $$

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