What is the integral of (-9x+3)^4 ?

The integral of (-9x+3)^4 is -1/45 (-9x+3)^5+C

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integral of (-9x+3)^4

\int (- 9 x + 3)^{4} d x

- \frac{1}{45} (- 9 x + 3)^{5} + C

Solution steps

\int (- 9 x + 3)^{4} d x

Apply u-substitution

: \int - \frac{u ^{4}}{9} d u

= \int - \frac{u ^{4}}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= - \frac{1}{9} \cdot \frac{u ^{5}}{5}

Substitute back

u = - 9 x + 3

= - \frac{1}{9} \cdot \frac{( - 9 x + 3 ) ^{5}}{5}

Simplify

- \frac{1}{9} \cdot \frac{( - 9 x + 3 ) ^{5}}{5} : - \frac{1}{45} (- 9 x + 3)^{5}

= - \frac{1}{45} (- 9 x + 3)^{5}

Add a constant to the solution

= - \frac{1}{45} (- 9 x + 3)^{5} + C

30 e^{x} + x^{20}

\frac{d}{d x} (5 cos^{3} (x))

\int \frac{x ^{2} + x + 3}{x ^{4} + 6 x ^{2} + 9} d x

\int \frac{1}{- x ^{2}} d x

f (x) = \frac{e ^{x}}{x}, (1, e)

What is the integral of (-9x+3)^4 ?
The integral of (-9x+3)^4 is -1/45 (-9x+3)^5+C