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

The integral of (4+3x)^4 is 1/15 (4+3x)^5+C

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integral of (4+3x)^4

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

\frac{1}{15} (4 + 3 x)^{5} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

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

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = 4 + 3 x

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

Simplify

\frac{1}{3} \cdot \frac{( 4 + 3 x ) ^{5}}{5} : \frac{1}{15} (4 + 3 x)^{5}

= \frac{1}{15} (4 + 3 x)^{5}

Add a constant to the solution

= \frac{1}{15} (4 + 3 x)^{5} + C

n = 1 \sum \infty \frac{n - 1}{2 n + 1}

cos (x^{- 9})

\frac{d}{d x} (3 - \frac{7}{2 x ^{\frac{1}{8}}})

x^{2} + y^{2} = 4, x^{2} + y^{2} = 4 x

\frac{d}{d y} (\frac{y ^{\frac{3}{2}}}{5})