What is the integral of (x^3+2x)^5(6x^2+4) ?

The integral of (x^3+2x)^5(6x^2+4) is 1/3 (x^3+2x)^6+C

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integral of (x^3+2x)^5(6x^2+4)

\int (x^{3} + 2 x)^{5} (6 x^{2} + 4) d x

\frac{1}{3} (x^{3} + 2 x)^{6} + C

Solution steps

\int (x^{3} + 2 x)^{5} (6 x^{2} + 4) d x

Apply u-substitution

: \int 2 u^{5} d u

= \int 2 u^{5} d u

Take the constant out:

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

= 2 \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{3} + 2 x

= 2 \cdot \frac{( x ^{3} + 2 x ) ^{6}}{6}

Simplify

2 \cdot \frac{( x ^{3} + 2 x ) ^{6}}{6} : \frac{1}{3} (x^{3} + 2 x)^{6}

= \frac{1}{3} (x^{3} + 2 x)^{6}

Add a constant to the solution

= \frac{1}{3} (x^{3} + 2 x)^{6} + C

\frac{d}{d x} (ln (\frac{x}{2 e}))

f (x) = x^{2} + 9, (- 2, 13)

\int_{0}^{2} \frac{1}{x - 1} d x

\int (\frac{x}{x + 1}) d x

\frac{\partial}{\partial x} (y - 1)

What is the integral of (x^3+2x)^5(6x^2+4) ?
The integral of (x^3+2x)^5(6x^2+4) is 1/3 (x^3+2x)^6+C