What is the integral of e^{x(t-1)} ?

The integral of e^{x(t-1)} is 1/(-1+t)e^{(-1+t)x}+C

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integral of e^{x(t-1)}

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

\frac{1}{- 1 + t} e^{(- 1 + t) x} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = (- 1 + t) x

= \frac{1}{- 1 + t} e^{(- 1 + t) x}

Add a constant to the solution

= \frac{1}{- 1 + t} e^{(- 1 + t) x} + C

\frac{\partial}{\partial x} (arctan (x))

\frac{1}{s ^{2} + s ( \frac{3}{2} ) 2}

\int \frac{1}{7 x} d x

\int \frac{x ^{2}}{x ^{2} + x - 2} d x

\int \frac{5 x ^{3} - 3 x ^{2} + 2 x - 1}{x ^{4} + x ^{2}} d x

What is the integral of e^{x(t-1)} ?
The integral of e^{x(t-1)} is 1/(-1+t)e^{(-1+t)x}+C