What is the integral of (4x^2)/(sqrt(x^3+16)) ?

The integral of (4x^2)/(sqrt(x^3+16)) is 8/3 sqrt(x^3+16)+C

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integral of (4x^2)/(sqrt(x^3+16))

\int \frac{4 x ^{2}}{x ^{3} + 16} d x

\frac{8}{3} x^{3} + 16 + C

Solution steps

\int \frac{4 x ^{2}}{x ^{3} + 16} 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} + 16} 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} + 16

= 4 \cdot \frac{2}{3} x^{3} + 16

Simplify

4 \cdot \frac{2}{3} x^{3} + 16 : \frac{8}{3} x^{3} + 16

= \frac{8}{3} x^{3} + 16

Add a constant to the solution

= \frac{8}{3} x^{3} + 16 + C

y = x^{\frac{5}{3}} - x^{\frac{2}{3}}

x \to - 4 lim (x (x + 3) (x + 2))

\frac{x}{- 6 x + 6}

3 sin (x), 3 cos (x)

y^{^{'}} = t^{2} y^{3}

What is the integral of (4x^2)/(sqrt(x^3+16)) ?
The integral of (4x^2)/(sqrt(x^3+16)) is 8/3 sqrt(x^3+16)+C