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

The integral of (e^x+2x)^2 is 1/2 e^{2x}+4(e^xx-e^x)+(4x^3)/3+C

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

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

\frac{1}{2} e^{2 x} + 4 (e^{x} x - e^{x}) + \frac{4 x ^{3}}{3} + C

Solution steps

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

Expand

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

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

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

\int 4 e^{x} x d x = 4 (e^{x} x - e^{x})

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

= \frac{1}{2} e^{2 x} + 4 (e^{x} x - e^{x}) + \frac{4 x ^{3}}{3}

Add a constant to the solution

= \frac{1}{2} e^{2 x} + 4 (e^{x} x - e^{x}) + \frac{4 x ^{3}}{3} + C

\int \frac{x ^{3}}{1 - x ^{4}} d x

f (x) = 5 - 3 x

\int_{- 1}^{1} (x^{2} - x) d x

\int 4 r d r

\int \frac{1}{x - 9} d x

What is the integral of (e^x+2x)^2 ?
The integral of (e^x+2x)^2 is 1/2 e^{2x}+4(e^xx-e^x)+(4x^3)/3+C