What is the integral of (3x+1)^4 ?

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

Popular >

integral of (3x+1)^4

\int (3 x + 1)^{4} d x

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

Solution steps

\int (3 x + 1)^{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 = 3 x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d a} (sin (2 a) + sin (a))

\int (\frac{18 x ^{6} + 3 x ^{4}}{6 x ^{2}}) d x

\int_{0}^{\frac{π}{2}} 42

y = x^{3} + 4 x - 3, (1, 2)