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Question Number 68598 by Abdo msup. last updated on 14/Sep/19

$$find\int \frac{dx}{x^3-4x+3}$$

Commented by turbo msup by abdo last updated on 14/Sep/19

$$letI=\int \frac{dx}{x^3-4x+3}letdecompose$$$$F\left(x\right)=\frac{1}{x^3-4x+3}wehave$$$$x^3-4x+3=x^3-x-3x+3$$$$=x(x^2-1)-3(x-1)$$$$=(x-1)(x(x+1)-3)$$$$=(x-1)(x^2+x-3)\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{bx+c}{x^2+x-3}=\frac{1}{(x-1)(x^2+x-3)}$$$$a={lim}_{x\to 1}(x-1)F\left(x\right)=-1$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=1$$$$\Rightarrow F\left(x\right)=\frac{-1}{x-1}+\frac{x+c}{x^2+x-3}$$$$F\left(0\right)=\frac{1}{3}=1-\frac{c}{3}\Rightarrow 1=3-c\Rightarrow$$$$c=2\Rightarrow F\left(x\right)=-\frac{1}{x-1}+\frac{x+2}{x^2+x-3}$$$$\Rightarrow I=\int (-\frac{1}{x-1}+\frac{x+2}{x^2+x-3})dx$$$$=-ln\mid x-1\mid +\frac{1}{2}\int \frac{2x+1+3}{x^2+x-3}dx$$$$=-ln\mid x-1\mid +\frac{1}{2}ln\mid x^2+x-3\mid$$$$+\frac{3}{2}\int \frac{dx}{x^2+x-3}wehave$$$$\int \frac{dx}{x^2+x-3}=\int \frac{dx}{x^2+2x\frac{1}{2}+\frac{1}{4}-3-\frac{1}{4}}$$$$=\int \frac{dx}{{(x+\frac{1}{2})}^2-\frac{13}{4}}$$$${=}_{x+\frac{1}{2}=\frac{\sqrt{13}}{2}u}\frac{4}{13}\int \frac{1}{u^2-1}\frac{\sqrt{13}}{2}du$$$$=\frac{1}{\sqrt{13}}\int (\frac{1}{u-1}-\frac{1}{u+1})du$$$$=\frac{1}{\sqrt{13}}ln\mid \frac{u-1}{u+1}\mid +c$$$$=\frac{1}{\sqrt{13}}ln\mid \frac{\frac{2x+1}{\sqrt{13}}-1}{\frac{2x+1}{\sqrt{13}}+1}\mid \Rightarrow$$$$I=-ln\mid x-1\mid +\frac{1}{2}ln\mid x^2+x-3\mid$$$$+\frac{3}{2\sqrt{13}}ln\mid \frac{2x+1-\sqrt{13}}{2x+1+\sqrt{13}}\mid +c.$$$$$$$$$$

Commented by som(math1967) last updated on 14/Sep/19

$$\int \frac{dx}{(x-1)(x^2+x-3)}$$$$let(x-1)(Ax+B)+C(x^2+x-3)=1$$$$puttingx=1.....C=-1$$$$x=0........B=2$$$$x=-1........A=1$$$$\therefore \int \frac{\{(x-1)(x+2)-(x^2+x-3)\}dx}{(x-1)(x^2+x-3)}$$$$=\int \frac{(x+2)dx}{(x^2+x-3)}-\int \frac{dx}{(x-1)}$$$$=\frac{1}{2}\int \frac{(2x+4)dx}{x^2+x-3}-ln(x-1)$$$$=\frac{1}{2}\int \frac{d(x^2+x-3)}{(x^2+x-3)}+\frac{3}{2}\int \frac{dx}{{(x+\frac{1}{2})}^2-{\left(\frac{\sqrt{13}}{2}\right)}^2}-lnx$$$$=\frac{1}{2}ln(x^2+x-3)+\frac{3}{2}\times \frac{1}{\sqrt{13}}ln\left(\frac{x+\frac{1}{2}-\frac{\sqrt{13}}{2}}{x+\frac{1}{2}+\frac{\sqrt{13}}{2}}\right)-ln(x-1)+K$$

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