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

The integral of e^{0.2x} is 5e^{0.2x}+C

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integral of e^{0.2x}

\int e^{0.2 x} d x

5 e^{0.2 x} + C

Solution steps

\int e^{0.2 x} d x

Apply u-substitution

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

= \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

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

Simplify

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

Use the common integral:

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

= 5 e^{u}

Substitute back

u = 0.2 x

= 5 e^{0.2 x}

Add a constant to the solution

= 5 e^{0.2 x} + C

y^{^{''}} - 9 y^{^{'}} = 81 t, y (0) = 9, y^{^{'}} (0) = 0

\int \frac{- 3}{7 e ^{7 x}} d x

\frac{d}{d x} (\frac{2}{e ^{x} - e ^{- x}})

y = 4 x, y = 3 x^{2}

x \frac{d y}{d x} - 3 y = 6