What is the integral of ae^{ax}(2+ax) ?

The integral of ae^{ax}(2+ax) is ae^{ax}x+e^{ax}+C

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integral of ae^{ax}(2+ax)

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

a e^{a x} x + e^{a x} + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= a \cdot \int 2 e^{a x} + a e^{a x} 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

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

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

\int a e^{a x} x d x = e^{a x} x - \frac{e ^{a x}}{a}

= a (e^{a x} \frac{2}{a} + e^{a x} x - \frac{e ^{a x}}{a})

Simplify

a (e^{a x} \frac{2}{a} + e^{a x} x - \frac{e ^{a x}}{a}) : a e^{a x} x + e^{a x}

= a e^{a x} x + e^{a x}

Add a constant to the solution

= a e^{a x} x + e^{a x} + C

\frac{d y}{d x} = sin (19 x)

\int_{4}^{5} t t - 4 d t

y = 6 x - 2

\frac{\partial}{\partial x} (arctan (x \cdot t))

2^{t}

What is the integral of ae^{ax}(2+ax) ?
The integral of ae^{ax}(2+ax) is ae^{ax}x+e^{ax}+C