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Mila [183]
4 years ago
15

Evaluate the integral x4 – 2x2 + 4x + 1 x3 – x2 - x + 1 dx

Mathematics
1 answer:
hjlf4 years ago
3 0

Looks like the integral is

\displaystyle\int\frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}\,\mathrm dx

First simplify the integrand by long division (or however you like):

\dfrac{x^4-2x^2+4x+1}{x^3-x^2-x+1}=x+1+\dfrac{4x}{x^3-x^2-x+1}

Also notice that

x^3-x^2-x+1=x^2(x-1)-(x-1)=(x^2-1)(x-1)=(x+1)(x-1)^2

Split the last term into partial fractions:

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

4x=a(x-1)^2+b(x+1)(x-1)+c(x+1)

4x=(a+b)x^2+(c-2a)x+a-b+c

\implies\begin{cases}a+b=0\\c-2a=4\\a-b+c=0\end{cases}\implies a=-1,b=1,c=2

So the integral is equivalent to

\displaystyle\int\left(x+1-\frac1{x+1}+\frac1{x-1}+\frac2{(x-1)^2}\right)\,\mathrm dx

=\displaystyle\frac{x^2}2+x-\ln|x+1|-\ln|x-1|-\frac2{x-1}+C

=\displaystyle\frac{x^2}2+x-\ln|x^2-1|-\frac2{x-1}+C

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