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

The integral of (x+1)e^{4x+1} is e(1/4 e^{4x}x+3/16 e^{4x})+C

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

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

e (\frac{1}{4} e^{4 x} x + \frac{3}{16} e^{4 x}) + C

Solution steps

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

(x + 1) e^{4 x + 1} = e^{1} (x + 1) e^{4 x}

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

Simplify

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

Apply Integration By Parts

: \frac{1}{4} e^{4 x} (x + 1) - \int \frac{1}{4} e^{4 x} d x

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

\int \frac{1}{4} e^{4 x} d x = \frac{1}{16} e^{4 x}

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

Expand

\frac{1}{4} e^{4 x} (x + 1) - \frac{1}{16} e^{4 x} : \frac{1}{4} e^{4 x} x + \frac{3}{16} e^{4 x}

= e (\frac{1}{4} e^{4 x} x + \frac{3}{16} e^{4 x})

Add a constant to the solution

= e (\frac{1}{4} e^{4 x} x + \frac{3}{16} e^{4 x}) + C

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What is the integral of (x+1)e^{4x+1} ?
The integral of (x+1)e^{4x+1} is e(1/4 e^{4x}x+3/16 e^{4x})+C