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Question Number 71802 by mathmax by abdo last updated on 20/Oct/19

$$find\int \frac{x^2-1}{{(x+3)}^2(x^3-5x+4)}dx$$

Commented by mathmax by abdo last updated on 20/Oct/19

$$letI=\int \frac{x^2-1}{{(x+3)}^2(x^3-5x+4)}firstwehave$$$$x^3-5x+4=x^3-1-5x+5=(x-1)(x^2+x+1)-5(x-1)$$$$=(x-1)(x^2+x+1-5)=(x-1)(x^2+x-4)\Rightarrow$$$$I=\int \frac{(x-1)(x+1)}{{(x+3)}^2(x-1)(x^2+x-4)}dx=\int \frac{dx}{{(x+3)}^2(x^2+x-4)}$$$$rootsof{x}^2+x-4\to \mathrm{\Delta }=1-4(-4)=17\Rightarrow x_1=\frac{-1+\sqrt{17}}{2}$$$$and{x}_2=\frac{-1-\sqrt{17}}{2}letdecomposeF\left(x\right)=\frac{1}{{(x+3)}^2(x^2+x-4)}$$$$=\frac{1}{{(x+3)}^2(x-x_1)(x-x_2)}\Rightarrow F\left(x\right)=\frac{a}{x+3}+\frac{b}{{(x+3)}^2}+\frac{c}{x-x_1}+\frac{d}{x-x_2}$$$$b=\frac{1}{(-3-x_1)(-3-x_2)}=\frac{1}{(x_1+3)(x_2+3)}$$$$=\frac{1}{(\frac{-1+\sqrt{17}}{2}+3)(\frac{-1-\sqrt{17}}{2}+3)}=\frac{4}{(5+\sqrt{17})(5-\sqrt{17})}=\frac{4}{25-17}$$$$=\frac{4}{8}=\frac{1}{2}$$$$c=\frac{1}{{(x_1+3)}^2(x_1-x_2)}=\frac{1}{{(\frac{-1+\sqrt{17}}{2}+3)}^2\left(\sqrt{17}\right)}=\frac{4}{\sqrt{17}{(5+\sqrt{17})}^2}$$$$d=\frac{1}{{(x_2+3)}^2(x_2-x_1)}=\frac{1}{{(\frac{-1-\sqrt{17}}{2}+3)}^2(-\sqrt{17})}=\frac{-4}{\sqrt{17}{(5-\sqrt{17})}^2}$$$$F\left(0\right)=-\frac{1}{36}=\frac{a}{3}+\frac{b}{9}-\frac{c}{x_1}-\frac{d}{x_2}\Rightarrow -\frac{1}{12}=a+\frac{b}{3}-\frac{3c}{x_1}-\frac{3d}{x_2}\Rightarrow$$$$a=-\frac{1}{12}-\frac{b}{3}+\frac{3c}{x_1}+\frac{3d}{x_2}\Rightarrow I=\int F\left(x\right)dx$$$$I=aln\mid x+3\mid -\frac{b}{x+3}+cln\mid x-x_1\mid +dln\mid x-x_2\mid +C$$

Answered by MJS last updated on 20/Oct/19

$$\int \frac{(x-1)(x+1)}{(x-1){(x+3)}^2(x^2+x-4)}dx=$$$$=\int \frac{x+1}{{(x+3)}^2(x^2+x-4)}dx=$$$$=-\int \frac{dx}{{(x+3)}^2}-2\int \frac{dx}{x+3}+(1+\frac{4}{\sqrt{17}})\int \frac{dx}{x+\frac{1+\sqrt{17}}{2}}+(1-\frac{4}{\sqrt{17}})\int \frac{dx}{x+\frac{1-\sqrt{17}}{2}}=$$$$=\frac{1}{x+3}-2\ln \mid x+3\mid +(1+\frac{4}{\sqrt{17}})\ln \mid x+\frac{1+\sqrt{17}}{2}\mid +(1-\frac{4}{\sqrt{17}})\ln \mid x+\frac{1-\sqrt{17}}{2}\mid +C$$

Commented by mathmax by abdo last updated on 20/Oct/19

$$thnkssir.$$

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