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

The integral of 1/(z^2(z^2+1)) is -1/z-arctan(z)+C

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integral of 1/(z^2(z^2+1))

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

- \frac{1}{z} - arctan (z) + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{z} - arctan (z)

Add a constant to the solution

= - \frac{1}{z} - arctan (z) + C

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What is the integral of 1/(z^2(z^2+1)) ?
The integral of 1/(z^2(z^2+1)) is -1/z-arctan(z)+C