What is the integral of (2x+6)^5 ?

The integral of (2x+6)^5 is 1/12 (2x+6)^6+C

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integral of (2x+6)^5

\int (2 x + 6)^{5} d x

\frac{1}{12} (2 x + 6)^{6} + C

Solution steps

\int (2 x + 6)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

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

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 2 x + 6

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

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 6 ) ^{6}}{6} : \frac{1}{12} (2 x + 6)^{6}

= \frac{1}{12} (2 x + 6)^{6}

Add a constant to the solution

= \frac{1}{12} (2 x + 6)^{6} + C

\int_{2}^{\infty} \frac{7}{3 x ^{2}} d x

\int \frac{1}{x ( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

3 cos^{3} (t)

(ln (y))^{6} \frac{d y}{d x} = x^{6} y, y (1) = e^{2}

y = e^{x} (3 - 5 x + 3 x^{2})