What is the integral of (x^2)/(sqrt(7+3x^3)) ?

The integral of (x^2)/(sqrt(7+3x^3)) is 2/9 sqrt(7+3x^3)+C

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integral of (x^2)/(sqrt(7+3x^3))

\int \frac{x ^{2}}{7 + 3 x ^{3}} d x

\frac{2}{9} 7 + 3 x^{3} + C

Solution steps

\int \frac{x ^{2}}{7 + 3 x ^{3}} d x

Apply u-substitution

: \int \frac{u}{2 x 7 + 3 u ^{\frac{3}{2}}} d u

= \int \frac{1}{3 7 + 3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{7 + 3 u} d u

Apply u-substitution

: \int \frac{1}{3 v} d v

= \frac{1}{3} \cdot \int \frac{1}{3 v} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= \frac{1}{3} \cdot \frac{1}{3} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} \cdot 2 (7 + 3 x^{3})^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{3} \cdot 2 (7 + 3 x^{3})^{\frac{1}{2}} : \frac{2}{9} 7 + 3 x^{3}

= \frac{2}{9} 7 + 3 x^{3}

Add a constant to the solution

= \frac{2}{9} 7 + 3 x^{3} + C

\frac{\partial}{\partial x} (e^{14 x y})

\frac{d}{d x} (3 x^{2} - 18 x + 24)

x \cdot sec (x)

n = 0 \sum \infty 4 + (- 1)^{n}

\int_{0}^{\infty} e^{- x - y} d x

What is the integral of (x^2)/(sqrt(7+3x^3)) ?
The integral of (x^2)/(sqrt(7+3x^3)) is 2/9 sqrt(7+3x^3)+C