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

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

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integral of (1-x)^b

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

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

Solution steps

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

Apply u-substitution

: \int - u^{b} d u

= \int - u^{b} d u

Take the constant out:

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

= - \int u^{b} d u

Apply the Power Rule

: \frac{u ^{b + 1}}{b + 1}

= - \frac{u ^{b + 1}}{b + 1}

Substitute back

u = 1 - x

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

Add a constant to the solution

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

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

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

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

\int x^{3} + 3 x^{2} d x

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