What is the integral of (ax+b)^n ?

The integral of (ax+b)^n is ((ax+b)^{n+1})/(a(n+1))+C

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integral of (ax+b)^n

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

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

Solution steps

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

Apply u-substitution

= \int \frac{u ^{n}}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int u^{n} d u

Apply the Power Rule

= \frac{1}{a} \cdot \frac{u ^{n + 1}}{n + 1}

Substitute back

u = a x + b

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

Simplify

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

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

Add a constant to the solution

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

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