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

The integral of (3x+2)(3x^2+4x)^4 is 1/10 (3x^2+4x)^5+C

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

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

\frac{1}{10} (3 x^{2} + 4 x)^{5} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = 3 x^{2} + 4 x

= \frac{1}{2} \cdot \frac{( 3 x ^{2} + 4 x ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot \frac{( 3 x ^{2} + 4 x ) ^{5}}{5} : \frac{1}{10} (3 x^{2} + 4 x)^{5}

= \frac{1}{10} (3 x^{2} + 4 x)^{5}

Add a constant to the solution

= \frac{1}{10} (3 x^{2} + 4 x)^{5} + C

x \to - \infty lim (5^{x})

ln (\frac{( x + 2 )}{x - 2})

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

\int \frac{( x ^{2} )}{16 - x ^{2}} d x

\frac{d}{d x} (x^{2} h)

What is the integral of (3x+2)(3x^2+4x)^4 ?
The integral of (3x+2)(3x^2+4x)^4 is 1/10 (3x^2+4x)^5+C