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

The integral of (4x^2)/(sqrt(x^3)+8) is 8/3 (x^{3/2}+8-8ln|x^{3/2}+8|)+C

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integral of (4x^2)/(sqrt(x^3)+8)

\int \frac{4 x ^{2}}{x ^{3} + 8} d x

\frac{8}{3} (x^{\frac{3}{2}} + 8 - 8 ln x^{\frac{3}{2}} + 8) + C

Solution steps

\int \frac{4 x ^{2}}{x ^{3} + 8} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{x ^{3} + 8} d x

Apply u-substitution

: \int \frac{2 ( u - 8 )}{3 u} d u

= 4 \cdot \int \frac{2 ( u - 8 )}{3 u} d u

Take the constant out:

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

= 4 \cdot \frac{2}{3} \cdot \int \frac{u - 8}{u} d u

Expand

\frac{u - 8}{u} : 1 - \frac{8}{u}

= 4 \cdot \frac{2}{3} \cdot \int 1 - \frac{8}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{2}{3} (\int 1 d u - \int \frac{8}{u} d u)

\int 1 d u = u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 4 \cdot \frac{2}{3} (u - 8 ln ∣ u ∣)

Substitute back

u = x^{3} + 8

= 4 \cdot \frac{2}{3} (x^{3} + 8 - 8 ln x^{3} + 8)

Simplify

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

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

Add a constant to the solution

= \frac{8}{3} (x^{\frac{3}{2}} + 8 - 8 ln x^{\frac{3}{2}} + 8) + C

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What is the integral of (4x^2)/(sqrt(x^3)+8) ?
The integral of (4x^2)/(sqrt(x^3)+8) is 8/3 (x^{3/2}+8-8ln|x^{3/2}+8|)+C