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

The integral of (3x+2)^5 is 1/18 (3x+2)^6+C

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integral of (3x+2)^5

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

\frac{1}{18} (3 x + 2)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= \int \frac{u ^{5}}{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 u^{5} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 3 x + 2

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

Simplify

\frac{1}{3} \cdot \frac{( 3 x + 2 ) ^{6}}{6} : \frac{1}{18} (3 x + 2)^{6}

= \frac{1}{18} (3 x + 2)^{6}

Add a constant to the solution

= \frac{1}{18} (3 x + 2)^{6} + C

\frac{d}{d x} (e^{2 x^{4}})

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

\frac{\partial}{\partial x} (8 x (x^{2} + y^{2})^{- \frac{1}{2}})

f (x) = x^{2} + 13, at x = 6

\int 5^{1 - x} d x