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

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

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

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

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

Solution steps

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

(1 - x) e^{1 + 2 x} = e^{1} (1 - x) e^{2 x}

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

Take the constant out:

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

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

Simplify

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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