What is the integral of (z^4)/((z^2-1)^2) ?

The integral of (z^4)/((z^2-1)^2) is z-3/4 ln|z+1|-1/(4(z+1))+3/4 ln|z-1|-1/(4(z-1))+C

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integral of (z^4)/((z^2-1)^2)

\int \frac{z ^{4}}{( z ^{2} - 1 ) ^{2}} d z

z - \frac{3}{4} ln ∣ z + 1 ∣ - \frac{1}{4 ( z + 1 )} + \frac{3}{4} ln ∣ z - 1 ∣ - \frac{1}{4 ( z - 1 )} + C

Solution steps

\int \frac{z ^{4}}{( z ^{2} - 1 ) ^{2}} d z

Take the partial fraction of

\frac{z ^{4}}{( z ^{2} - 1 ) ^{2}} : 1 - \frac{3}{4 ( z + 1 )} + \frac{1}{4 ( z + 1 ) ^{2}} + \frac{3}{4 ( z - 1 )} + \frac{1}{4 ( z - 1 ) ^{2}}

= \int 1 - \frac{3}{4 ( z + 1 )} + \frac{1}{4 ( z + 1 ) ^{2}} + \frac{3}{4 ( z - 1 )} + \frac{1}{4 ( z - 1 ) ^{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 1 d z - \int \frac{3}{4 ( z + 1 )} d z + \int \frac{1}{4 ( z + 1 ) ^{2}} d z + \int \frac{3}{4 ( z - 1 )} d z + \int \frac{1}{4 ( z - 1 ) ^{2}} d z

\int 1 d z = z

\int \frac{3}{4 ( z + 1 )} d z = \frac{3}{4} ln ∣ z + 1 ∣

\int \frac{1}{4 ( z + 1 ) ^{2}} d z = - \frac{1}{4 ( z + 1 )}

\int \frac{3}{4 ( z - 1 )} d z = \frac{3}{4} ln ∣ z - 1 ∣

\int \frac{1}{4 ( z - 1 ) ^{2}} d z = - \frac{1}{4 ( z - 1 )}

= z - \frac{3}{4} ln ∣ z + 1 ∣ - \frac{1}{4 ( z + 1 )} + \frac{3}{4} ln ∣ z - 1 ∣ - \frac{1}{4 ( z - 1 )}

Add a constant to the solution

= z - \frac{3}{4} ln ∣ z + 1 ∣ - \frac{1}{4 ( z + 1 )} + \frac{3}{4} ln ∣ z - 1 ∣ - \frac{1}{4 ( z - 1 )} + C

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What is the integral of (z^4)/((z^2-1)^2) ?
The integral of (z^4)/((z^2-1)^2) is z-3/4 ln|z+1|-1/(4(z+1))+3/4 ln|z-1|-1/(4(z-1))+C