What is the integral of (a+bx)^k ?

The integral of (a+bx)^k is ((a+bx)^{k+1})/(b(k+1))+C

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integral of (a+bx)^k

\int (a + b x)^{k} d x

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

Solution steps

\int (a + b x)^{k} d x

Apply u-substitution

: \int \frac{u ^{k}}{b} d u

= \int \frac{u ^{k}}{b} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int u^{k} d u

Apply the Power Rule

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

= \frac{1}{b} \cdot \frac{u ^{k + 1}}{k + 1}

Substitute back

u = a + b x

= \frac{1}{b} \cdot \frac{( a + b x ) ^{k + 1}}{k + 1}

Simplify

\frac{1}{b} \cdot \frac{( a + b x ) ^{k + 1}}{k + 1} : \frac{( a + b x ) ^{k + 1}}{b ( k + 1 )}

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

Add a constant to the solution

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

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What is the integral of (a+bx)^k ?
The integral of (a+bx)^k is ((a+bx)^{k+1})/(b(k+1))+C