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Question Number 50384 by prof Abdo imad last updated on 16/Dec/18

$$find\int \frac{dx}{(1-x^2)(1-x^3)}$$$$2)calculate{\int }_2^{\sqrt{5}}\frac{dx}{(1-x^2)(1-x^3)}$$

Commented by Abdo msup. last updated on 23/Dec/18

$$letdecomposeF\left(x\right)=\frac{1}{(1-x^2)(1-x^3)}$$$$F\left(x\right)=\frac{1}{{(x-1)}^2(x+1)(x^2+x+1)}$$$$=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{x+1}+\frac{dx+f}{x^2+x+1}$$$$b={lim}_{x\to 1}{(x-1)}^{2}F\left(x\right)=\frac{1}{6}$$$$c={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{4}\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{6{(x-1)}^2}+\frac{1}{4(x+1)}+\frac{dx+f}{x^2+x+1}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=a+\frac{1}{4}+d=0\Rightarrow a+d=-\frac{1}{4}\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{6{(x-1)}^2}+\frac{1}{4(x+1)}+\frac{(-a-\frac{1}{4})x+f}{x^2+x+1}$$$$F\left(0\right)=1=-a+\frac{1}{6}+\frac{1}{4}+f\Rightarrow -a+f=1-\frac{1}{6}-\frac{1}{4}$$$$=\frac{5}{6}-\frac{1}{4}=\frac{14}{24}=\frac{7}{12}$$$$F\left(2\right)=a+\frac{1}{6}+\frac{1}{12}+\frac{(-2a-\frac{1}{2})+f}{7}=\frac{1}{21}\Rightarrow$$$$(1-\frac{2}{7})a+\frac{1}{4}-\frac{1}{14}+\frac{f}{7}=\frac{1}{21}\Rightarrow$$$$\frac{5}{7}a+\frac{10}{56}+\frac{f}{7}=\frac{1}{21}\Rightarrow 5a+\frac{70}{56}+f=\frac{1}{3}\Rightarrow$$$$5a+\frac{35}{28}+f=\frac{1}{3}\Rightarrow 5a+f=\frac{1}{3}-\frac{35}{28}$$$$becontinued.....$$$$$$

Answered by ajfour last updated on 16/Dec/18

$$I=\int \frac{dx}{(1-x^2)(1-x^3)}$$$$Let\frac{1}{{(1-x)}^2(1+x)(1+x+x^2)}$$$$=\frac{A}{{(1-x)}^2}+\frac{B}{1-x}+\frac{C}{1+x}+\frac{Dx+E}{1+x+x^2}$$$$A=\frac{1}{6},C=\frac{1}{4}$$$$1=\frac{1}{6}(1+x)(1+x+x^2)$$$$+B(1+x)(1-x^3)$$$$+\frac{1}{4}(1-x)(1-x^3)$$$$+(Dx+E){(1-x)}^2(1+x)$$$$coeff.of{x}^4=-B+\frac{1}{4}+D=0$$$$coeff.of{x}^3=\frac{1}{6}-B-\frac{1}{4}+E-D=0$$$$constanttermcoeff.$$$$=\frac{1}{6}+B+\frac{1}{4}+E=1$$$$....$$

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