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

The integral of (4x^2)/(sqrt(x^3+8)) is 8/3 sqrt(x^3+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^{3} + 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}{3} d u

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

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{2}{3} u

Substitute back

u = x^{3} + 8

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

Simplify

4 \cdot \frac{2}{3} x^{3} + 8 : \frac{8}{3} x^{3} + 8

= \frac{8}{3} x^{3} + 8

Add a constant to the solution

= \frac{8}{3} x^{3} + 8 + C

\frac{\partial}{\partial x} ((1 - x - 4 y)^{\frac{1}{2}})

\frac{\partial}{\partial x} (5 e^{x^{5} y})

f (x) = \frac{x ^{2} + 2 x + 1}{x ^{2} - 2 x + 1}

\int 7 sin (5 x) sin (x) d x

x \to 0 lim (4 x - ∣ x ∣)

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