Can anyone help me with this indefinite integral?

Question

Couldn't solve this indefinite integral, can someone help me? $$\int \frac {x^3+4x^2+6x+1}{x^3+x^2+x-3} dx$$

Answer 1

First, do long division to change your improper rational expression into a polynomial plus a proper rational expression. $$\int 1 +\frac{3x^2+5x+4}{x^3+x^2+x-3} \; dx.$$ Then factor the denominator by noting that if you plug in $x=1$ you get $0$, so $x-1$ is a factor. Then do partial fractions to get $$\int 1 + \frac{2}{x-1} + \frac{x+2}{x^2+2x+3} \; dx.$$ Finally, complete the square in the quadratic to get $$\frac{(x+1)+1}{(x+1)^2+2}$$ and substitute $u=x+1$. That leaves $$\int 1 \;dx + \int \frac{2}{x-1} \; dx + \int \frac{u}{u^2+2} \; du + \int \frac{1}{u^2+2} \; du.$$

Answer 2

$$x^3+4x^2+6x+1=(x^3+x^2+x-3)+3x^2+5x+4$$ Thus, $$\int\frac{x^3+4x^2+6x+1}{x^3+x^2+x-3}dx=\int\left(1+\frac{3x^2+5x+4}{x^3+x^2+x-3}\right)dx=\int\left(1+\frac{3x^2+5x+4}{(x-1)(x^2+2x+3)}\right)dx$$

Let $$\frac{3x^2+5x+4}{(x-1)(x^2+2x+3)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2x+3}$$

Find A,B, and C and solve.