What is the integral of (1-x)^9 ?

The integral of (1-x)^9 is -((1-x)^{10})/(10)+C

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integral of (1-x)^9

\int (1 - x)^{9} d x

- \frac{( 1 - x ) ^{10}}{10} + C

Solution steps

\int (1 - x)^{9} d x

Apply u-substitution

: \int - u^{9} d u

= \int - u^{9} d u

Take the constant out:

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

= - \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= - \frac{u ^{10}}{10}

Substitute back

u = 1 - x

= - \frac{( 1 - x ) ^{10}}{10}

Add a constant to the solution

= - \frac{( 1 - x ) ^{10}}{10} + C

f (x) = \frac{sin ( x )}{cos ( x )}

\int x^{2} - 2 d x

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

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

\frac{d}{d x} (a e^{- 5 x} + b e^{- 3 x} + c x e^{- x})

What is the integral of (1-x)^9 ?
The integral of (1-x)^9 is -((1-x)^{10})/(10)+C