integration with partial fractions

[lislr8]

Hi,

Int [(2x^3-7x^2+6x-21)/(x+1)(x-2)(x-3)]dx

I know for this question to use partial fractions, however, when I go about it the normal way I end up with C=0, which can't happen. I get A= -3, B=7, and C=0.

I heard from that you have to divide first but what do I divide and how would that change things?

Thanks!

 

[soroban]

Hello, lislr8!

\(\displaystyle \int \frac{2x^3-7x^2+6x-21}{(x+1)(x-2)(x-3)}\,dx\)

The numerator and denominator are both cubics; it is an "improper" fraction.

\(\displaystyle \text{Using long division, we have: }\;2 + \frac{x^2+8x - 33}{(x+1)(x-2)(x-3)}\)

. . \(\displaystyle \text{and that fraction reduces: }\;\frac{(x+11)(x-3)}{(x+1)(x-2)(x-3)} \:=\:\frac{x+11}{(x+1)(x-2)}\)

\(\displaystyle \text{And that's why }C = 0\text{; there is }no\text{ fraction with }(x-3)\text{ in the denominator.}\)


\(\displaystyle \text{Then we have: }\;\int\bigg[2 + \frac{x+11}{(x+1)(x-2)}\bigg]\,dx \;=\;\int\bigg[2 + \frac{-\frac{10}{3}}{x+1} + \frac{\frac{13}{3}}{x-2}\bigg]\,dx\)


. . . . . . . . \(\displaystyle =\;\int 2\,dx \;-\; \tfrac{10}{3}\!\1\int\frac{dx}{x+1} \;+\;\tfrac{13}{3}\!\!\int\frac{dx}{x-2}\)

Got it?