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

The integral of (1+2x)^2 is x+2x^2+(4x^3)/3+C

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integral of (1+2x)^2

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

x + 2 x^{2} + \frac{4 x ^{3}}{3} + C

Solution steps

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

Expand

(1 + 2 x)^{2} : 1 + 4 x + 4 x^{2}

= \int 1 + 4 x + 4 x^{2} d x

Apply the Sum Rule:

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

= \int 1 d x + \int 4 x d x + \int 4 x^{2} d x

\int 1 d x = x

\int 4 x d x = 2 x^{2}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

= x + 2 x^{2} + \frac{4 x ^{3}}{3}

Add a constant to the solution

= x + 2 x^{2} + \frac{4 x ^{3}}{3} + C

\frac{- 3 x - 2}{2 x ^{2} + x + 3}

\frac{d}{d x} (\frac{e ^{x}}{x ^{3}})

\frac{\partial}{\partial y} (- 2 x e^{x y})

y^{^{'}} = (x - 14)^{2}

y^{^{''}} + 324 y = sec (18 x), y (0) = 2, y^{^{'}} (0) = - 1

What is the integral of (1+2x)^2 ?
The integral of (1+2x)^2 is x+2x^2+(4x^3)/3+C