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

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

Popular >

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

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

- \frac{z ^{2}}{2} - ln - z^{2} + 1 + C

Solution steps

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

Long division

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

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

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

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

= - \frac{z ^{2}}{2} - ln - z^{2} + 1

Add a constant to the solution

= - \frac{z ^{2}}{2} - ln - z^{2} + 1 + C

\int \frac{7 x + 6}{x ^{2} + x + 7} d x

\int (4 x^{5} + 2 x^{4} + 3 x^{2} - x + 6) d x

\int sin^{6} (x) \cdot cos^{3} (x) d x

\int \frac{1}{ln ( u )} d u

\int e^{- 2 x} + 1 d x

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