What is the integral of e^{1-3x} ?

The integral of e^{1-3x} is -1/3 e^{1-3x}+C

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integral of e^{1-3x}

\int e^{1 - 3 x} d x

- \frac{1}{3} e^{1 - 3 x} + C

Solution steps

\int e^{1 - 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{3} e^{u}

Substitute back

u = 1 - 3 x

= - \frac{1}{3} e^{1 - 3 x}

Add a constant to the solution

= - \frac{1}{3} e^{1 - 3 x} + C

x \to \infty lim (x^{1})

y^{^{''}} + 2 y = e x + x^{3}

\int 3 x^{2} + 3 d x

x \to 9 + lim (\frac{1}{x - 9})