What is the integral of (z^2+2z+3)/((z-6)(z^2+4)) ?

The integral of (z^2+2z+3)/((z-6)(z^2+4)) is 51/40 ln|z-6|+1/40 (-11/2 ln|z^2+4|+7arctan(z/2))+C

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integral of (z^2+2z+3)/((z-6)(z^2+4))

\int \frac{z ^{2} + 2 z + 3}{( z - 6 ) ( z ^{2} + 4 )} d z

\frac{51}{40} ln ∣ z - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln z^{2} + 4 + 7 arctan (\frac{z}{2})) + C

Solution steps

\int \frac{z ^{2} + 2 z + 3}{( z - 6 ) ( z ^{2} + 4 )} d z

Take the partial fraction of

\frac{z ^{2} + 2 z + 3}{( z - 6 ) ( z ^{2} + 4 )} : \frac{51}{40 ( z - 6 )} + \frac{- 11 z + 14}{40 ( z ^{2} + 4 )}

= \int \frac{51}{40 ( z - 6 )} + \frac{- 11 z + 14}{40 ( z ^{2} + 4 )} d z

Apply the Sum Rule:

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

= \int \frac{51}{40 ( z - 6 )} d z + \int \frac{- 11 z + 14}{40 ( z ^{2} + 4 )} d z

\int \frac{51}{40 ( z - 6 )} d z = \frac{51}{40} ln ∣ z - 6 ∣

\int \frac{- 11 z + 14}{40 ( z ^{2} + 4 )} d z = \frac{1}{40} (- \frac{11}{2} ln z^{2} + 4 + 7 arctan (\frac{z}{2}))

= \frac{51}{40} ln ∣ z - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln z^{2} + 4 + 7 arctan (\frac{z}{2}))

Add a constant to the solution

= \frac{51}{40} ln ∣ z - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln z^{2} + 4 + 7 arctan (\frac{z}{2})) + C

What is the integral of (z^2+2z+3)/((z-6)(z^2+4)) ?
The integral of (z^2+2z+3)/((z-6)(z^2+4)) is 51/40 ln|z-6|+1/40 (-11/2 ln|z^2+4|+7arctan(z/2))+C