What is the integral of x^2(3+4x^3)^9 ?

The integral of x^2(3+4x^3)^9 is 1/120 (3+4x^3)^{10}+C

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integral of x^2(3+4x^3)^9

\int x^{2} (3 + 4 x^{3})^{9} d x

\frac{1}{120} (3 + 4 x^{3})^{10} + C

Solution steps

\int x^{2} (3 + 4 x^{3})^{9} d x

Apply u-substitution

: \int \frac{u ( 3 + 4 u ^{\frac{3}{2}} ) ^{9}}{2 x} d u

= \int \frac{( 3 + 4 u ) ^{9}}{3} 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 (3 + 4 u)^{9} d u

Apply u-substitution

: \int \frac{v ^{9}}{4} d v

= \frac{1}{3} \cdot \int \frac{v ^{9}}{4} 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}{4} \cdot \int v^{9} d v

Apply the Power Rule

: \frac{v ^{10}}{10}

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{v ^{10}}{10}

Substitute back

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 3 + 4 x ^{3} ) ^{10}}{10}

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 3 + 4 x ^{3} ) ^{10}}{10} : \frac{1}{120} (3 + 4 x^{3})^{10}

= \frac{1}{120} (3 + 4 x^{3})^{10}

Add a constant to the solution

= \frac{1}{120} (3 + 4 x^{3})^{10} + C

x \to o lim (f (x, \frac{1}{x}))

x \to 5 lim (x - 2)

x \to 0 lim (\frac{x ^{2}}{e ^{x}})

\frac{\partial}{\partial z} (yz e^{x z})

\frac{d y}{d x} = \frac{3 x ^{2} - 1}{2 y}

What is the integral of x^2(3+4x^3)^9 ?
The integral of x^2(3+4x^3)^9 is 1/120 (3+4x^3)^{10}+C