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

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

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

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

\frac{2}{3} x^{3} + 4 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{3} d u

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

Integral of a constant:

\int a d x = a x

= \frac{2}{3} u

Substitute back

u = x^{3} + 4

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

Add a constant to the solution

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

10 y^{^{''}} - 7 y^{^{'}} - 12 y = 0

x \to 2 - lim (x + 3)

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (1) = - 2, y^{^{'}} (- 2) = - 3

\int \frac{x}{1 + x ^{3}} d x

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 20 x e^{x}

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