Question

Integration by Parts Fractions

1.) ∫dz/z+z^3

2.) ∫(2s+1)ds/s^2(s^2+1)

Answer 1

1) $\intop \frac{dz}{z+z^3}=\intop \frac{dz}{z(1+z^2)}=\intop \frac{(z^2+1-z^2)dz}{z(1+z^2)}=\intop (\frac{z^2+1}{z(1+z^2)}-\frac{z^2}{z(1+z^2)})dz=$

$$=\int(\frac{1}{z}-\frac{z}{1+z^2})dz =\int \frac{1}{z}dz-\int\frac{z}{1+z^2}dz=ln|z|-\frac{1}{2}\int\frac{2z}{1+z^2}dz=$$ $$=ln|z|-\frac{1}{2}\int \frac{d(1+z^2)}{1+z^2}=ln |z|-\frac{1}{2}ln(1+z^2)+C$$

2) Use partial fractions:

$$\frac{2s+1}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1}=\frac{As^3+As+Bs^2+B+Cs^3+Ds^2}{s^2(s^2+1)}$$ $$2s+1=(A+C)s^3+(B+D)s^2+As+B$$ $$\begin{cases} B=1, \\ A=2, \\ B+D=0, \\ A+C=0 \end{cases}$$ $$\begin{cases} B=1, \\ A=2, \\ D=-1, \\ C=-2 \end{cases}$$ $$\frac{2s+1}{s^2(s^2+1)}=\frac{2}{s}+\frac{1}{s^2}-\frac{2s+1}{s^2+1}$$

Hense,

$$\int \frac{(2s+1)ds}{s^2(s^2+1)}=\int\frac{2}{s}ds+\int\frac{1}{s^2}ds-\int\frac{2s+1}{s^2+1}ds=$$

$$=2ln|s|-\frac{1}{s}-\int\frac{2s}{s^2+1}ds-\int\frac{ds}{s^2+1}=$$ $$=2ln|s|-\frac{1}{s}-ln(s^2+1)-tan^{-1}(s^2+1)+C$$