What is the integral of z(1+i) ?

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

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integral of z(1+i)

integral

z (1 + i)

\frac{z ^{2}}{2} + i \frac{z ^{2}}{2} + C

Solution steps

\int z (1 + i) d z

Take the constant out:

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

= (1 + i) \cdot \int z d z

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

= (1 + i) \frac{z ^{2}}{2}

Simplify

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

= \frac{z ^{2}}{2} + i \frac{z ^{2}}{2}

Add a constant to the solution

= \frac{z ^{2}}{2} + i \frac{z ^{2}}{2} + C

r^{- 1}

ln ((\frac{1}{x}) - 1)

\frac{( x ^{2} - a ) ^{3}}{2}

t (1 - t)

\frac{1}{( 4 + x ^{2} )}