What is the integral of 4(6x+1)^5 ?

The integral of 4(6x+1)^5 is 1/9 (6x+1)^6+C

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integral of 4(6x+1)^5

\int 4 (6 x + 1)^{5} d x

\frac{1}{9} (6 x + 1)^{6} + C

Solution steps

\int 4 (6 x + 1)^{5} d x

Take the constant out:

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

= 4 \cdot \int (6 x + 1)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{6} d u

= 4 \cdot \int \frac{u ^{5}}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int u^{5} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{6} \cdot \frac{u ^{6}}{6}

Substitute back

u = 6 x + 1

= 4 \cdot \frac{1}{6} \cdot \frac{( 6 x + 1 ) ^{6}}{6}

Simplify

4 \cdot \frac{1}{6} \cdot \frac{( 6 x + 1 ) ^{6}}{6} : \frac{1}{9} (6 x + 1)^{6}

= \frac{1}{9} (6 x + 1)^{6}

Add a constant to the solution

= \frac{1}{9} (6 x + 1)^{6} + C

f (x) = 3 x^{3} + 3 x^{2} + 3 x

\int_{0}^{1} x^{4} e^{8 x} d x

f (x) = (x - 2) (2 x + 3)

(1 - x^{2}) y^{^{''}} + 2 x y^{^{'}} - 2 y = 0

f (x) = ln (x^{3} - 5 x^{2} + 3 x - 1)