Answer
$$\int\frac{dt}{t^3+t^2-2t}=\frac{1}{3}\ln|t-1|+\frac{1}{6}\ln|t+2|-\frac{1}{2}\ln|t|+C$$
Work Step by Step
$$A=\int\frac{dt}{t^3+t^2-2t}$$
1) Express the integrand as a sum of partial fractions:
$$\frac{1}{t^3+t^2-2t}=\frac{1}{t(t^2+t-2)}=\frac{1}{t(t-1)(t+2)}$$ $$=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+2}$$
Clear fractions:
$$1=A(t^2+t-2)+B(t^2+2t)+C(t^2-t)$$ $$1=At^2+At-2A+Bt^2+2Bt+Ct^2-Ct$$ $$1=(A+B+C)t^2+(A+2B-C)t-2A$$
Equating coefficients of corresponding powers of $t$, we get
- $-2A=1$
- $A+B+C=0$
- $A+2B-C=0$
Calculation gives us $A=-1/2$, $B=1/3$ and $C=1/6$
Therefore, $$\frac{1}{t^3+t^2-2t}=\frac{1}{3(t-1)}+\frac{1}{6(t+2)}-\frac{1}{2t}$$
2) Evaluate the integral: $$A=\frac{1}{3}\int\frac{dt}{t-1}+\frac{1}{6}\int\frac{dt}{t+2}-\frac{1}{2}\int\frac{dt}{t}$$ $$A=\frac{1}{3}\ln|t-1|+\frac{1}{6}\ln|t+2|-\frac{1}{2}\ln|t|+C$$
