# Is it possible to solve following integral with partial-fraction decomposition?

I have a integral $$\int_0^1\frac{-x^2+4x+4}{x^2-4}~dx$$ Which I changed to $$\int_0^1\frac{-x^2+4x+4}{(x-2)(x+2)}~dx$$ But I don't know how to change numerator to have lesser polynom degree than the denominator to use partial-fraction decomposition.

• Use long division first – Jean-Sébastien Jan 17 '16 at 17:05
• Start by dividing $x^2-4$ into $-x^2+4x+4$. – Tim Raczkowski Jan 17 '16 at 17:05

Big Hint: \begin{align} \int_0^1\frac{-x^2+4x+4}{x^2-4}\,\mathrm{d}x &=\int_0^1\left(\frac{4x}{x^2-4}-1\right)\mathrm{d}x\\ &=\int_0^1\left(\frac2{x-2}+\frac2{x+2}-1\right)\mathrm{d}x\\ \end{align}

First divide numerator by denominator, then apply partial fraction decomposition to the resulting fraction from the remainder divided by the denominator (and integrate the quotient part separately).

The numerator is

$$8-(x-2)^2$$

so the integrand is

\begin{align}\frac{8-(x-2)^2}{(x-2)(x+2)} &= \frac{8}{(x-2)(x+2)} - \frac{x-2}{x+2} \\ &= 2 \left (\frac1{x-2} - \frac1{x+2} \right ) - \frac{4}{x+2} +1 \end{align}

You should be able to go from here...