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

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

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integral of (n-1)/n e^{(n-1)x}

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

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

Solution steps

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

Take the constant out:

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

= \frac{n - 1}{n} \cdot \int e^{(n - 1) x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{n - 1}{n} \cdot \frac{1}{n - 1} e^{u}

Substitute back

u = (n - 1) x

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

Simplify

\frac{n - 1}{n} \cdot \frac{1}{n - 1} e^{(n - 1) x} : \frac{e ^{x (n - 1)}}{n}

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

Add a constant to the solution

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

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