What is the integral of (0.1q+1)e^{0.3q} ?

The integral of (0.1q+1)e^{0.3q} is 0.33333…e^{0.3q}q+2.22222…e^{0.3q}+C

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integral of (0.1q+1)e^{0.3q}

\int (0.1 q + 1) e^{0.3 q} d q

0.33333 \dots e^{0.3 q} q + 2.22222 \dots e^{0.3 q} + C

Solution steps

\int (0.1 q + 1) e^{0.3 q} d q

Expand

(0.1 q + 1) e^{0.3 q} : 0.1 e^{0.3 q} q + e^{0.3 q}

= \int 0.1 e^{0.3 q} q + e^{0.3 q} d q

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 0.1 e^{0.3 q} q d q + \int e^{0.3 q} d q

\int 0.1 e^{0.3 q} q d q = 0.1 (3.33333 \dots e^{0.3 q} q - 11.11111 \dots e^{0.3 q})

\int e^{0.3 q} d q = 3.33333 \dots e^{0.3 q}

= 0.1 (3.33333 \dots e^{0.3 q} q - 11.11111 \dots e^{0.3 q}) + 3.33333 \dots e^{0.3 q}

Simplify

0.1 (3.33333 \dots e^{0.3 q} q - 11.11111 \dots e^{0.3 q}) + 3.33333 \dots e^{0.3 q} : 0.33333 \dots e^{0.3 q} q + 2.22222 \dots e^{0.3 q}

= 0.33333 \dots e^{0.3 q} q + 2.22222 \dots e^{0.3 q}

Add a constant to the solution

= 0.33333 \dots e^{0.3 q} q + 2.22222 \dots e^{0.3 q} + C

What is the integral of (0.1q+1)e^{0.3q} ?
The integral of (0.1q+1)e^{0.3q} is 0.33333…e^{0.3q}q+2.22222…e^{0.3q}+C