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Evaluate \(\int \frac{4 x}{x^{3}-x^{2}-x+1} d x\)

(1) \(x=-1 \rightarrow F(-1)=0\)

(2) \(x+1\) معامل الحدودية

(3) \(\frac{x^{3}-x^{2}-x+1}{x+1}\)

\(\begin{array}{r}{\frac{x^{2}-2 x+1}{x^{3}-x^{2}-x+1}} \\ {\frac{x^{3}+x^{2}}{0-2 x^{2}-x+1}=} \\ {\frac{-2 x^{2}-2 x}{0+x+1}} \\ {\frac{+x+1}{2}}\end{array}\)

\(x^{2}-2 x+1\)

\(\frac{x^{3}-x^{2}-x+1}{x+1}=x^{2}-2 x+1\)

\(x^{3}-x^{2}-x+1=\left(x^{2}-2 x+1\right)(x+1)\)

\(x^{2}-2 x+1\)

\(\begin{array}{l}{a=1} \\ {b=-2} \\ {c=+1}\end{array}\)

\(x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=1\)

\((x-1)(x-1)=(x-1)^{2}\)

\(x^{3}-x^{2}-x+1=(x-1)^{2}(x+1)\)

(1) \(f(x)=\frac{4 x}{x^{3}-x^{2}-x+1}=\frac{4 x}{(x+1)(x-1)^{2}}\)

\(=\frac{A}{x+1}+\frac{B}{(x-1)^{2}}+\frac{c}{x-1}\)

\(A=(x+1) f(x)\left|=(x+1) \frac{4 x}{(x+1)(x-1)^{2}}\right| = -1\)

\(\begin{array}{c}{B=(x-1)^{2} \cdot f(x)\left|=(x-1)^{2} \cdot \frac{4 x}{(x+1)(x-1)^{2}}\right|=2} \\ {x = 1}\end{array}\)

\(\begin{aligned} c=(x-1) f(x) &\left|(x-1) \cdot \frac{4 x}{(x+1)(x-1)^{2}}\right|=\\ & x=1 \end{aligned}\)

\(\frac{4 (1)}{2(0)}\)

\(f(x)=\frac{A}{x+1}+\frac{B}{(x-1)^{2}}+\frac{c}{x-1}\)

\(\frac{4 x}{(x+1)(x-1)^{2}}=\frac{-1}{(x+1)}+\frac{2}{(x-1)^{2}}+\frac{c}{(x-1)}\)

\(\frac{u x}{(x+1)(x-1)^{2}}=\frac{-1(x-1)^{2}}{(x+1)(x-1)^{2}}+\frac{2(x+1)}{(x-1)^{2}(x+1)}+\frac{c(x-1)(x+1)}{(x-1)(x-1)(x+1)}\)

\(4 x=-(x-1)^{2}+2(x+1)+c(x-1)(x+1)\)

\(4 x=-(x-1)^{2}+2(x+1)+c\left(x^{2}-1\right)\)

\(0=c-1 \Rightarrow [c=1] \quad[A=-1] \quad [B=2]\)

\(f(x)=\frac{-1}{x+1}+\frac{2}{(x-1)^{2}}+\frac{1}{(x-1)}\)

\(\int f(x)=\int \frac{-1}{x+1} d x+\int \frac{2}{(x-1)^{2}}+\int \frac{1}{(x-1)} d x\)

\(\int f(x)=-\ln |x+1|+\int \frac{2}{(x-1)^{2}} d x+\ln |x-1|\)

\(* \int \frac{2}{(x-1)^{2}} d x=2 \int \frac{1}{(x-1)^{2}} d x\)

let \(u=x-1 \rightarrow d u=1 d x\)

\(=2 \int \frac{1}{u^{2}} d u=2 \int u^{-2} d u=2\left[\frac{u^{-1}}{-1}\right]=-2 \frac{1}{u}\)

\(\int \frac{4}{(x+1)(x-1)^{2}}=-\ln |x+1|-2 \frac{1}{x-1}+\ln |x-1|+c\)

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