What is the integral of e^{rt} ?

The integral of e^{rt} is 1/r e^{rt}+C

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integral of e^{rt}

\int e^{r t} d t

\frac{1}{r} e^{r t} + C

Solution steps

\int e^{r t} d t

Apply u-substitution

: \int \frac{e ^{u}}{r} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{r} e^{u}

Substitute back

u = r t

= \frac{1}{r} e^{r t}

Add a constant to the solution

= \frac{1}{r} e^{r t} + C

\int 2 x e^{x^{2} + y^{2}} d x

x \to 2 + lim (\frac{1}{x - 2} - \frac{1}{x - 2})

\int \frac{1}{u ( 1 - u )} d u

x^{3} + 1, - x + 1, y = 0

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