What is the integral of (2x+9)^5 ?

The integral of (2x+9)^5 is 1/12 (2x+9)^6+C

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integral of (2x+9)^5

\int (2 x + 9)^{5} d x

\frac{1}{12} (2 x + 9)^{6} + C

Solution steps

\int (2 x + 9)^{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 + 9

= \frac{1}{2} \cdot \frac{( 2 x + 9 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 9 ) ^{6}}{6} : \frac{1}{12} (2 x + 9)^{6}

= \frac{1}{12} (2 x + 9)^{6}

Add a constant to the solution

= \frac{1}{12} (2 x + 9)^{6} + C

f (x) = \frac{4 x ^{2}}{x + 3}, (1, 1)

y = \frac{x}{x + 1}, (4, 0.4)

y^{^{''}} + 4 y = 5 csc^{2} (2 t)

\frac{\partial}{\partial z} (y sin (z))

\frac{tan ( x ) - 5}{sec ( x )}