Integration by parts/partial fractions

[nameVoid]

\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

 

[gabbagabbahey]

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}|

 

[nameVoid]

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

 

[gabbagabbahey]

<br /> (x-\frac{1}{2})ln(x^2-x+2)-2x+1+\frac{7}{\sqrt{7}}tan^-1\frac{2x-1}{\sqrt{7}}<br />

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