Integrate ((z^2+3z-1)/(z+1))dz?

Question

if possible, don't solve it by partial fraction.

Answer 1

Long division: you should get:

(z + 2) - 3/(z+1)

Integrating gives you:
½z² + 2z - ln|(z+1)³| + C

Answer 2

It's not exactly partial fractions. If you divide z^2+3z-1 by z+1, you get

z^2+3z-1 = (z +1)(z+2) - 3, so that, for z <>1,

(z^2+3z-1)/(z+1) = z +2 - 3/(z +1). It follows that

Integral ((z^2+3z-1)/(z+1))dz = z^2/2 + 2z - 3 ln(|z +1|) + C, C constant.

Answer 3

(z^2+3z-1)/(z+1)
= (z^2 + z + 2z + 2 - 3) / (z + 1)
= [ z(z + 1) + 2(z + 1) - 3] / (z + 1)
= z + 2 - 3/(z + 1)

∫[(z^2+3z-1)/(z+1)] dz
= ∫ z dz + 2 ∫ dz - ∫ 3/(z + 1) dz
= z^2/2 + 2z - 3 ln l z + 1 l + c

Earlier I had posted last term as +ln l z + 1 l, but realized my mistake after seeing Matt's first step and have editted my answer.

Answer 4

I concur. Division gives you z + 2 -3/(z+1). This integrates to (1/2)z^2 + 2z - 3ln|z+1| + k (some constant).