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

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

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

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

- \frac{z ^{3}}{3} + z + C

Solution steps

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

Factor

\frac{1 - z ^{4}}{z ^{2} + 1}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Cancel the common factor:

z^{2} + 1

= - \int (z + 1) (z - 1) d z

Expand

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

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

\int z^{2} d z = \frac{z ^{3}}{3}

\int 1 d z = z

= - (\frac{z ^{3}}{3} - z)

Simplify

= - \frac{z ^{3}}{3} + z

Add a constant to the solution

= - \frac{z ^{3}}{3} + z + C

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