What is the integral of (8m^2)/((m^3+2)^3) ?

The integral of (8m^2)/((m^3+2)^3) is -4/(3(m^3+2)^2)+C

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integral of (8m^2)/((m^3+2)^3)

\int \frac{8 m ^{2}}{( m ^{3} + 2 ) ^{3}} d m

- \frac{4}{3 ( m ^{3} + 2 ) ^{2}} + C

Solution steps

\int \frac{8 m ^{2}}{( m ^{3} + 2 ) ^{3}} d m

Take the constant out:

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

= 8 \cdot \int \frac{m ^{2}}{( m ^{3} + 2 ) ^{3}} d m

Apply u-substitution

: \int \frac{1}{3 u ^{3}} d u

= 8 \cdot \int \frac{1}{3 u ^{3}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 8 \cdot \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = m^{3} + 2

= 8 \cdot \frac{1}{3} (- \frac{1}{2 ( m ^{3} + 2 ) ^{2}})

Simplify

8 \cdot \frac{1}{3} (- \frac{1}{2 ( m ^{3} + 2 ) ^{2}}) : - \frac{4}{3 ( m ^{3} + 2 ) ^{2}}

= - \frac{4}{3 ( m ^{3} + 2 ) ^{2}}

Add a constant to the solution

= - \frac{4}{3 ( m ^{3} + 2 ) ^{2}} + C

x \to 0 + lim (x^{\frac{5}{x}})

x \to 0 lim (\frac{3}{10 - e ^{x}})

\int 2 e^{2}

\frac{d}{d x} (x lo g_{2} (x))

n = 1 \sum \infty - \frac{1}{8 ^{n}}

What is the integral of (8m^2)/((m^3+2)^3) ?
The integral of (8m^2)/((m^3+2)^3) is -4/(3(m^3+2)^2)+C