# Integration by parts/partial fractions

$$\int ln(x^2-x+2)dx$$
$$\int ln( (x-\frac{1}{2})^2+\frac{7}{4} )dx$$
$$u=x-\frac{1}{2}$$
$$\int ln(u^2+\frac{7}{4})du$$
$$uln(u^2+\frac{7}{4})-\int\frac{2u^2}{u^2+\frac{7}{4}}du$$
$$uln(u^2+\frac{7}{4})-ln|u^2+\frac{7}{4}|+C$$
$$(x-\frac{1}{2})ln(x^2-x+2)-ln|x^2-x+2|+C$$

## Answers and Replies

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gabbagabbahey
Homework Helper
Gold Member
$$uln(u^2+\frac{7}{4})-\int\frac{2u^2}{u^2+\frac{7}{4}}du$$
$$uln(u^2+\frac{7}{4})-ln|u^2+\frac{7}{4}|+C$$
Not quite,

$$\int\frac{2u^2}{u^2+\frac{7}{4}}du\neq \ln|u^2+\frac{7}{4}|$$

eh,
$$-\int \frac{2u^2}{u^2+\frac{7}{4}}du=\int 2-\frac{7}{2(u^2+\frac{7}{4})}du$$
$$-2u+\frac{7}{\sqrt{7}}tan^{-1}\frac{2u}{\sqrt{7}}+C$$
$$(x-\frac{1}{2})ln(x^2-x+2)-2x+1+\frac{7}{\sqrt{7}}tan^-1\frac{2x-1}{\sqrt{7}}$$

gabbagabbahey
Homework Helper
Gold Member
$$(x-\frac{1}{2})ln(x^2-x+2)-2x+1+\frac{7}{\sqrt{7}}tan^-1\frac{2x-1}{\sqrt{7}}$$
That looks more or less correct. You are, of course, missing a constant of integration here, and since 1 is also a constant you might as well absorb it into the integration constant and write it as

$$\int \ln\left(x^2-x+2\right)dx=\left(x-\frac{1}{2}\right)\ln\left(x^2-x+2\right)-2x+\sqrt{7}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)+C$$