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 - \frac{1}{0.1} e^{u} d u

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

x \to 1 lim (ln (x^{\frac{1}{2}}))

\frac{1}{s ^{3} + 5 s}

\frac{1}{1 + s a}

\frac{d}{d x} f (x) = 8^{(x^{2} + 4 x)}

\frac{d y}{d x} = e^{2 x + 3 y}, y (0) = 0