What is the integral of (m^3)/((z^2)^3) ?

The integral of (m^3)/((z^2)^3) is (m^4)/(4z^6)+C

integral of (m^3)/((z^2)^3)

\int \frac{m ^{3}}{( z ^{2} ) ^{3}} d m

\frac{m ^{4}}{4 z ^{6}} + C

Solution steps

\int \frac{m ^{3}}{( z ^{2} ) ^{3}} d m

Take the constant out:

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

= \frac{1}{( z ^{2} ) ^{3}} \cdot \int m^{3} d m

(z^{2})^{3} = z^{6}

= \frac{1}{z ^{6}} \cdot \int m^{3} d m

Apply the Power Rule

= \frac{1}{z ^{6}} \cdot \frac{m ^{4}}{4}

Simplify

\frac{1}{z ^{6}} \cdot \frac{m ^{4}}{4} : \frac{m ^{4}}{4 z ^{6}}

= \frac{m ^{4}}{4 z ^{6}}

Add a constant to the solution

= \frac{m ^{4}}{4 z ^{6}} + C

\frac{\partial}{\partial y} (x e^{x y})

\int \frac{( ln ( x ) ) ^{3}}{3 x} d x

n = 0 \sum \infty \frac{n x ^{n}}{n + 8}

\int \frac{3 x}{2 x ^{2} + 3 x + 4} d x

\frac{\partial}{\partial x} (2 x^{6 y})

What is the integral of (m^3)/((z^2)^3) ?
The integral of (m^3)/((z^2)^3) is (m^4)/(4z^6)+C