What is the integral of (x+1)(x^2+2x) ?

The integral of (x+1)(x^2+2x) is 1/4 (x^2+2x)^2+C

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integral of (x+1)(x^2+2x)

\int (x + 1) (x^{2} + 2 x) d x

\frac{1}{4} (x^{2} + 2 x)^{2} + C

Solution steps

\int (x + 1) (x^{2} + 2 x) d x

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = x^{2} + 2 x

= \frac{1}{2} \cdot \frac{( x ^{2} + 2 x ) ^{2}}{2}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 2 x ) ^{2}}{2} : \frac{1}{4} (x^{2} + 2 x)^{2}

= \frac{1}{4} (x^{2} + 2 x)^{2}

Add a constant to the solution

= \frac{1}{4} (x^{2} + 2 x)^{2} + C

\frac{d}{d x} (\frac{( x ^{4} )}{8} + \frac{1}{4 x ^{2}})

x \to - 3 lim (\frac{2}{x ^{2} - 9})

x y^{^{'}} = 2 x^{2} + y

\frac{d y}{d x} = 3 y^{2}

\int \frac{- 1}{y} d y

What is the integral of (x+1)(x^2+2x) ?
The integral of (x+1)(x^2+2x) is 1/4 (x^2+2x)^2+C