What is the integral of e^{0.1x} ?

The integral of e^{0.1x} is 10e^{0.1x}+C

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integral of e^{0.1x}

\int e^{0.1 x} d x

10 e^{0.1 x} + C

Solution steps

\int e^{0.1 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.1} d u

= \int e^{u} \frac{1}{0.1} d u

Take the constant out:

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

= \frac{1}{0.1} \cdot \int e^{u} d u

Simplify

= 10 \cdot \int e^{u} d u

Use the common integral:

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

= 10 e^{u}

Substitute back

u = 0.1 x

= 10 e^{0.1 x}

Add a constant to the solution

= 10 e^{0.1 x} + C

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