What is the integral of (z^2)/(z+4) ?

The integral of (z^2)/(z+4) is ((z+4)^2)/2-8(z+4)+16ln|z+4|+C

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integral of (z^2)/(z+4)

\int \frac{z ^{2}}{z + 4} d z

\frac{( z + 4 ) ^{2}}{2} - 8 (z + 4) + 16 ln ∣ z + 4 ∣ + C

Solution steps

\int \frac{z ^{2}}{z + 4} d z

Apply u-substitution

: \int \frac{( u - 4 ) ^{2}}{u} d u

= \int \frac{( u - 4 ) ^{2}}{u} d u

Expand

\frac{( u - 4 ) ^{2}}{u} : u - 8 + \frac{16}{u}

= \int u - 8 + \frac{16}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u - \int 8 d u + \int \frac{16}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{2}}{2} - 8 u + 16 ln ∣ u ∣

Substitute back

u = z + 4

= \frac{( z + 4 ) ^{2}}{2} - 8 (z + 4) + 16 ln ∣ z + 4 ∣

Add a constant to the solution

= \frac{( z + 4 ) ^{2}}{2} - 8 (z + 4) + 16 ln ∣ z + 4 ∣ + C

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What is the integral of (z^2)/(z+4) ?
The integral of (z^2)/(z+4) is ((z+4)^2)/2-8(z+4)+16ln|z+4|+C