How to solve partial fractions in integrals

Question

How to solve partial fractions in integrals $$\int { \frac1{x^ 3-x^ 2} }\,\mathrm dx$$

I got the answer as $-\ln|x|+1/x+\ln|x-1|+C$

Can anyone tell whether this correct or not

Answer 1

First write it in form $$\frac{1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$$ Multiplying both sides by $x^2(x-1)$, then we get $$1=(A+C)x^2+(B-A)x-B$$ By comparing coefficients you should get that $$A=B=-1$$ and $$C=1$$ I hope you know about how to integrate after integration.

Answer 2

Hint: use that $$\frac{1}{x^3-x^2}=-{x}^{-2}-{x}^{-1}+ \left( x-1 \right) ^{-1}$$

Answer 3

The partial fraction decomposition of this rational function has the form $$\frac1{x^3-x^2}=\frac1{x^2(x-1)}=\frac Ax+\frac B{x^2}+\frac C{c-1}\qquad(A,B,C\in \mathbf R).$$

To obtain the values of $A, B$ and $C$, multiply both sides by the l.c.m. of the denominators: $$1=Ax(x-1)+B(x-1)+Cx^2$$ Setting $x=0$, then $x=1$, you obtain $B=-1$ and $C=1$. To obtain $A$, consider the coefficients of the leading monomial ($x^2$ on the right): $$0=A+C,\quad\text{whence }\;A=-1.$$