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

The integral of (z^2+4)^4z is 1/10 (z^2+4)^5+C

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integral of (z^2+4)^4z

\int (z^{2} + 4)^{4} z d z

\frac{1}{10} (z^{2} + 4)^{5} + C

Solution steps

\int (z^{2} + 4)^{4} z d z

Apply u-substitution

= \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = z^{2} + 4

= \frac{1}{2} \cdot \frac{( z ^{2} + 4 ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot \frac{( z ^{2} + 4 ) ^{5}}{5} : \frac{1}{10} (z^{2} + 4)^{5}

= \frac{1}{10} (z^{2} + 4)^{5}

Add a constant to the solution

= \frac{1}{10} (z^{2} + 4)^{5} + C

What is the integral of (z^2+4)^4z ?
The integral of (z^2+4)^4z is 1/10 (z^2+4)^5+C