What is the integral of 9e^{0.2x} ?

The integral of 9e^{0.2x} is 45e^{0.2x}+C

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integral of 9e^{0.2x}

\int 9 e^{0.2 x} d x

45 e^{0.2 x} + C

Solution steps

\int 9 e^{0.2 x} d x

Take the constant out:

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

= 9 \cdot \int e^{0.2 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.2} d u

= 9 \cdot \int e^{u} \frac{1}{0.2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{0.2} \cdot \int e^{u} d u

Simplify

= 9 \cdot 5 \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot 5 e^{u}

Substitute back

u = 0.2 x

= 9 \cdot 5 e^{0.2 x}

Simplify

= 45 e^{0.2 x}

Add a constant to the solution

= 45 e^{0.2 x} + C

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