What is the integral of (nx)(1-n)/n ?

The integral of (nx)(1-n)/n is (x^2(1-n))/2+C

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integral of (nx)(1-n)/n

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

\frac{x ^{2} ( 1 - n )}{2} + C

Solution steps

\int n x \frac{1 - n}{n} d x

Take the constant out:

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

= n \frac{1 - n}{n} \cdot \int x d x

Simplify

n \frac{1 - n}{n} : 1 - n

= (1 - n) \cdot \int x d x

\int x d x = \frac{x ^{2}}{2}

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Add a constant to the solution

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

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What is the integral of (nx)(1-n)/n ?
The integral of (nx)(1-n)/n is (x^2(1-n))/2+C