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

The integral of x(ax+b)^n is 1/(a^2)(((ax+b)^{n+2})/(n+2)-(b(ax+b)^{n+1})/(n+1))+C

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integral of x(ax+b)^n

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

\frac{1}{a ^{2}} (\frac{( a x + b ) ^{n + 2}}{n + 2} - \frac{b ( a x + b ) ^{n + 1}}{n + 1}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{n} ( u - b )}{a ^{2}} d u

= \int \frac{u ^{n} ( u - b )}{a ^{2}} d u

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int u^{n} (u - b) d u

Expand

u^{n} (u - b) : u^{1 + n} - b u^{n}

= \frac{1}{a ^{2}} \cdot \int u^{1 + n} - b u^{n} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a ^{2}} (\int u^{1 + n} d u - \int b u^{n} d u)

\int u^{1 + n} d u = \frac{u ^{n + 2}}{n + 2}

\int b u^{n} d u = \frac{b u ^{n + 1}}{n + 1}

= \frac{1}{a ^{2}} (\frac{u ^{n + 2}}{n + 2} - \frac{b u ^{n + 1}}{n + 1})

Substitute back

u = a x + b

= \frac{1}{a ^{2}} (\frac{( a x + b ) ^{n + 2}}{n + 2} - \frac{b ( a x + b ) ^{n + 1}}{n + 1})

Add a constant to the solution

= \frac{1}{a ^{2}} (\frac{( a x + b ) ^{n + 2}}{n + 2} - \frac{b ( a x + b ) ^{n + 1}}{n + 1}) + C

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What is the integral of x(ax+b)^n ?
The integral of x(ax+b)^n is 1/(a^2)(((ax+b)^{n+2})/(n+2)-(b(ax+b)^{n+1})/(n+1))+C