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

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

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

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

- \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{4} ln z^{2} + 1 + \frac{z}{2 ( 1 + z ^{2} )} + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{2 ( z + 1 )} d z = \frac{1}{2} ln ∣ z + 1 ∣

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

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

= - \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln z^{2} + 1 - arctan (z)) + \frac{1}{2} (arctan (z) + \frac{z}{1 + z ^{2}})

Simplify

- \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln z^{2} + 1 - arctan (z)) + \frac{1}{2} (arctan (z) + \frac{z}{1 + z ^{2}}) : - \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{4} ln z^{2} + 1 + \frac{z}{2 ( 1 + z ^{2} )}

= - \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{4} ln z^{2} + 1 + \frac{z}{2 ( 1 + z ^{2} )}

Add a constant to the solution

= - \frac{1}{2} ln ∣ z + 1 ∣ + \frac{1}{4} ln z^{2} + 1 + \frac{z}{2 ( 1 + z ^{2} )} + C

\int t^{2} t^{3} - 1 d t

\int_{0}^{π} sin (x) cos^{2} (x) d x

f (x) = \frac{x + 2}{7 x ^{2} e ^{x}}

\int_{0}^{t} e^{x} sin (t - x) d x

x \to 0 + lim (ln (2 x))

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