What is the integral of e^{ut} ?

The integral of e^{ut} is 1/u e^{ut}+C

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integral of e^{ut}

\int e^{u t} d t

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

Solution steps

\int e^{u t} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

v = u t

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

Add a constant to the solution

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

\frac{d y}{d x} = x y + 3 x - 2 y - 6

∣ X^{2} - 1 ∣

\int e^{t y} (- 4 y - 1) d y

(4 - \frac{1}{2 x ^{2}})^{- 1}

f (x) = x^{2} - 3 x - 4, (2, - 6)