What is the integral of 3(3x+9)^5 ?

The integral of 3(3x+9)^5 is 1/6 (3x+9)^6+C

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integral of 3(3x+9)^5

\int 3 (3 x + 9)^{5} d x

\frac{1}{6} (3 x + 9)^{6} + C

Solution steps

\int 3 (3 x + 9)^{5} d x

Take the constant out:

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

= 3 \cdot \int (3 x + 9)^{5} d x

Apply u-substitution

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

= 3 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x + 9

= 3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{6}}{6}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{6}}{6} : \frac{1}{6} (3 x + 9)^{6}

= \frac{1}{6} (3 x + 9)^{6}

Add a constant to the solution

= \frac{1}{6} (3 x + 9)^{6} + C

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 7, x (1) = 1

\int_{1}^{5} x^{2} - 6 x + 8 d x

\int 7 x^{4} - 5 x^{3} + 5 x^{2} + C d x

\frac{7}{x}, x = 1, x = 41