Partial fractions

1. Mar 15, 2010

nameVoid

$$\int \frac{2x+1}{4x^2+12x-7}dx$$
$$\frac{1}{4} \int \frac{2x+1}{x^2+3x-\frac{7}{4}}dx$$
$$\frac{1}{4} \int \frac{2x+1}{(x+\frac{3}{2})^2-4}dx$$
$$u=x+\frac{3}{2}$$
$$\frac{1}{2} \int \frac{u-1}{u^2-4}du$$
$$\frac{1}{2} \int \frac{u}{u^2-4}du -\frac{1}{2}\int \frac{du}{u^2-4}$$
$$\frac{1}{4} ln|u^2-4|-\frac{1}{2}\int \frac{A}{u+2} +\frac{B}{u-2} du$$
$$-\frac{1}{2}=A(u-2)+B(u+2)$$
$$A=\frac{1}{8}$$
$$B=-\frac{1}{8}$$
$$\frac{1}{4} ln|u^2-4|+\frac{1}{8}ln|\frac{u+2}{u-2}|+C$$
$$\frac{1}{4} ln|x^2+3x-\frac{7}{4}|-\frac{1}{8} ln|\frac{x+\frac{7}{2}}{x-\frac{1}{2}}|+C$$

2. Mar 15, 2010

gabbagabbahey

What you've got is correct, but it can still be simplified further; $\ln|u^2-4|=\ln|u-2|+\ln|u-2|$ and $\ln\left|\frac{u+2}{u-2}\right|=\ln|u-2|-\ln|u-2|$.

So,

$$\frac{1}{4}\ln|u^2-4|+\frac{1}{8}ln\left|\frac{u+2}{u-2}\right|=\frac{3}{8}\ln|u+2|+\frac{1}{8}\ln|u-2|$$

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