What is the integral of e^{0.34x+1} ?

The integral of e^{0.34x+1} is 2.94117…e^{0.34x+1}+C

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integral of e^{0.34x+1}

\int e^{0.34 x + 1} d x

2.94117 \dots e^{0.34 x + 1} + C

Solution steps

\int e^{0.34 x + 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.34} d u

= \int e^{u} \frac{1}{0.34} d u

Take the constant out:

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

= \frac{1}{0.34} \cdot \int e^{u} d u

Simplify

= 2.94117 \dots \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 2.94117 \dots e^{u}

Substitute back

u = 0.34 x + 1

= 2.94117 \dots e^{0.34 x + 1}

Add a constant to the solution

= 2.94117 \dots e^{0.34 x + 1} + C

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What is the integral of e^{0.34x+1} ?
The integral of e^{0.34x+1} is 2.94117…e^{0.34x+1}+C