What is the integral of (4x+1)^2e^{2x} ?

The integral of (4x+1)^2e^{2x} is 8e^{2x}x^2-4e^{2x}x+5/2 e^{2x}+C

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integral of (4x+1)^2e^{2x}

\int (4 x + 1)^{2} e^{2 x} d x

8 e^{2 x} x^{2} - 4 e^{2 x} x + \frac{5}{2} e^{2 x} + C

Solution steps

\int (4 x + 1)^{2} e^{2 x} d x

Expand

(4 x + 1)^{2} e^{2 x} : 16 e^{2 x} x^{2} + 8 e^{2 x} x + e^{2 x}

= \int 16 e^{2 x} x^{2} + 8 e^{2 x} x + e^{2 x} 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 16 e^{2 x} x^{2} d x + \int 8 e^{2 x} x d x + \int e^{2 x} d x

\int 16 e^{2 x} x^{2} d x = 16 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

\int 8 e^{2 x} x d x = 2 (2 e^{2 x} x - e^{2 x})

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

= 16 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + 2 (2 e^{2 x} x - e^{2 x}) + \frac{1}{2} e^{2 x}

Simplify

16 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + 2 (2 e^{2 x} x - e^{2 x}) + \frac{1}{2} e^{2 x} : 8 e^{2 x} x^{2} - 4 e^{2 x} x + \frac{5}{2} e^{2 x}

= 8 e^{2 x} x^{2} - 4 e^{2 x} x + \frac{5}{2} e^{2 x}

Add a constant to the solution

= 8 e^{2 x} x^{2} - 4 e^{2 x} x + \frac{5}{2} e^{2 x} + C

\int cos (\frac{2}{3} x) d x

y^{^{''}} + 4 y^{^{'}} + 3 y = 20 t e^{5 t}

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

\int \frac{x ^{3}}{4 + x ^{4}} d x

\frac{d}{d x} (7 x + 9)

What is the integral of (4x+1)^2e^{2x} ?
The integral of (4x+1)^2e^{2x} is 8e^{2x}x^2-4e^{2x}x+5/2 e^{2x}+C