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

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

(2) \(A=(x-1) f\left.(x)\right|_{x=1}= (x-1)\left.\frac{(4 x+1)}{(x+1)(x-1)}\right|_{x=1}=\left.\frac{4 x+1}{x+1}\right|_{x=1}=\frac{4(1)+1}{1+1}=2.5\)

\(B=(x+1) f\left.(x)\right|_{x=-1} =(x+1)\left.\frac{4 x+1}{(x+1)(x-1)}\right|_{x=-1} =\left.\frac{4 x+1}{x-1}\right|_{x=-1}=\frac{4(-1)+1}{-1-1}=1.5\)

(3) \(f(x)=\frac{2.5}{x-1}+\frac{1.5}{x+1}\)

(4) \(\int f(x) d x=\int \frac{4 x+1}{x^{2}-1} d x=\int \frac{2.5}{x-1}+\frac{1.5}{x+1} \cdot d x\)

\(=\int \frac{2.5}{x-1} d x+\int \frac{1.5}{x+1} d x\)

\(=2.5 \cdot \ln |x-1|+1.5 \ln |x+1|+C\)

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