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

The integral of (x^2-2x+1)e^{2x} is 1/2 e^{2x}x^2-3/2 e^{2x}x+5/4 e^{2x}+C

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

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

\frac{1}{2} e^{2 x} x^{2} - \frac{3}{2} e^{2 x} x + \frac{5}{4} e^{2 x} + C

Solution steps

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

Apply Integration By Parts

= \frac{1}{2} e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x} - \int e^{2 x} (x - 1) d x

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

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

Simplify

\frac{1}{2} e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x} - (\frac{1}{2} e^{2 x} x - \frac{3}{4} e^{2 x}) : \frac{1}{2} e^{2 x} x^{2} - \frac{3}{2} e^{2 x} x + \frac{5}{4} e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - \frac{3}{2} e^{2 x} x + \frac{5}{4} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{2} - \frac{3}{2} e^{2 x} x + \frac{5}{4} e^{2 x} + C

x \to π lim (2 cot (x))

\int e^{t^{3}} d t

\int 7 cos (4 x) d x

f (x) = 3 x^{4} - 7 ln (x)

\int 2 3 x d x

What is the integral of (x^2-2x+1)e^{2x} ?
The integral of (x^2-2x+1)e^{2x} is 1/2 e^{2x}x^2-3/2 e^{2x}x+5/4 e^{2x}+C