What is the integral of 4m(9m^2-10m) ?

The integral of 4m(9m^2-10m) is 9m^4-40/3 m^3+C

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integral of 4m(9m^2-10m)

\int 4 m (9 m^{2} - 10 m) d m

9 m^{4} - \frac{40}{3} m^{3} + C

Solution steps

\int 4 m (9 m^{2} - 10 m) d m

Take the constant out:

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

= 4 \cdot \int m (9 m^{2} - 10 m) d m

Multiply by the conjugate of

9 m^{2} - 10 m : \frac{81 m ^{4} - 100 m ^{2}}{9 m ^{2} + 10 m}

= 4 \cdot \int m \frac{81 m ^{4} - 100 m ^{2}}{9 m ^{2} + 10 m} d m

Simplify

m \frac{81 m ^{4} - 100 m ^{2}}{9 m ^{2} + 10 m} : m^{2} (9 m - 10)

= 4 \cdot \int m^{2} (9 m - 10) d m

Apply Integration By Parts

: \frac{1}{3} m^{3} (9 m - 10) - \int 3 m^{3} d m

= 4 (\frac{1}{3} m^{3} (9 m - 10) - \int 3 m^{3} d m)

\int 3 m^{3} d m = \frac{3 m ^{4}}{4}

= 4 (\frac{1}{3} m^{3} (9 m - 10) - \frac{3 m ^{4}}{4})

Simplify

4 (\frac{1}{3} m^{3} (9 m - 10) - \frac{3 m ^{4}}{4}) : 9 m^{4} - \frac{40}{3} m^{3}

= 9 m^{4} - \frac{40}{3} m^{3}

Add a constant to the solution

= 9 m^{4} - \frac{40}{3} m^{3} + C

ln (15 x)

f (x) = x^{12}

x \to - 3 lim (\frac{3 x + 1}{x - 3})

\frac{\partial}{\partial x} ((x^{5} + y^{3})^{7})

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

What is the integral of 4m(9m^2-10m) ?
The integral of 4m(9m^2-10m) is 9m^4-40/3 m^3+C