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

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

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integral of e^{tx-x}

\int e^{t x - x} d x

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

Solution steps

\int e^{t x - x} d x

Multiply by the conjugate of

- x + t x : \frac{t ^{2} x ^{2} - x ^{2}}{t x + x}

= \int e^{\frac{t ^{2} x ^{2} - x ^{2}}{t x + x}} d x

\frac{t ^{2} x ^{2} - x ^{2}}{t x + x} = x (t - 1)

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x (t - 1)

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

Add a constant to the solution

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

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