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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 = x^{3} + 2

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( ln ( x ) ) ^{4}}{x} d x

\int_{1}^{e} x^{2} ln (x) d x

\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

5 y^{^{''}} + 26 y = 0

e^{x^{2} + 9 x + 7}

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