∫5/x(x²+2x+5)dx
Is this integration by parts?
This will be a partial fractions problem.
Note that $\displaystyle x^2+2x+5$ is an irreducible quadratic.
Thus, $\displaystyle \frac{5}{x(x^2+2x+5)}=\frac{A}{x}+\frac{Bx+C}{x^2+ 2x+5}$
Thus, we see that $\displaystyle 5=A(x^2+2x+5)+Bx^2+Cx\implies 5=(A+B)x^2+(2A+C)x+5A$
Comparing coefficients, we have $\displaystyle A+B=0$, $\displaystyle 2A+C=0$ and $\displaystyle 5A=5\implies A=1$
Now, we see that $\displaystyle C=-2A=-2(1)=-2$ and $\displaystyle B=-A=-1$
So $\displaystyle \frac{5}{x(x^2+2x+5)}=\frac{1}{x}-\frac{x+2}{x^2+2x+5}$
Therefore, $\displaystyle \frac{5\,dx}{x(x^2+2x+5)}=\int\frac{\,dx}{x}-\int\frac{x+2}{x^2+2x+5}\,dx$
Can you take it from here?