What is the integral of (10z+2)/((z+1)^2(z-3)^2) ?

The integral of (10z+2)/((z+1)^2(z-3)^2) is 3/8 ln|z+1|+1/(2(z+1))-3/8 ln|z-3|-2/(z-3)+C

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integral of (10z+2)/((z+1)^2(z-3)^2)

\int \frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} d z

\frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3} + C

Solution steps

\int \frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} d z

Take the partial fraction of

\frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} : \frac{3}{8 ( z + 1 )} - \frac{1}{2 ( z + 1 ) ^{2}} - \frac{3}{8 ( z - 3 )} + \frac{2}{( z - 3 ) ^{2}}

= \int \frac{3}{8 ( z + 1 )} - \frac{1}{2 ( z + 1 ) ^{2}} - \frac{3}{8 ( z - 3 )} + \frac{2}{( z - 3 ) ^{2}} 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{3}{8 ( z + 1 )} d z - \int \frac{1}{2 ( z + 1 ) ^{2}} d z - \int \frac{3}{8 ( z - 3 )} d z + \int \frac{2}{( z - 3 ) ^{2}} d z

\int \frac{3}{8 ( z + 1 )} d z = \frac{3}{8} ln ∣ z + 1 ∣

\int \frac{1}{2 ( z + 1 ) ^{2}} d z = - \frac{1}{2 ( z + 1 )}

\int \frac{3}{8 ( z - 3 )} d z = \frac{3}{8} ln ∣ z - 3 ∣

\int \frac{2}{( z - 3 ) ^{2}} d z = - \frac{2}{z - 3}

= \frac{3}{8} ln ∣ z + 1 ∣ - (- \frac{1}{2 ( z + 1 )}) - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3}

Simplify

= \frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3}

Add a constant to the solution

= \frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3} + C

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What is the integral of (10z+2)/((z+1)^2(z-3)^2) ?
The integral of (10z+2)/((z+1)^2(z-3)^2) is 3/8 ln|z+1|+1/(2(z+1))-3/8 ln|z-3|-2/(z-3)+C