What is the integral of k/r (-(r^2)/4) ?

The integral of k/r (-(r^2)/4) is -(kr^2)/8+C

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integral of k/r (-(r^2)/4)

\int \frac{k}{r} (- \frac{r ^{2}}{4}) d r

- \frac{k r ^{2}}{8} + C

Solution steps

\int \frac{k}{r} (- \frac{r ^{2}}{4}) d r

Take the constant out:

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

= k \cdot \int \frac{1}{r} (- \frac{r ^{2}}{4}) d r

Simplify

\frac{1}{r} (- \frac{r ^{2}}{4}) : - \frac{r}{4}

= k \cdot \int - \frac{r}{4} d r

Take the constant out:

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

= k (- \frac{1}{4} \cdot \int r d r)

Apply the Power Rule

: \frac{r ^{2}}{2}

= k (- \frac{1}{4} \cdot \frac{r ^{2}}{2})

Simplify

k (- \frac{1}{4} \cdot \frac{r ^{2}}{2}) : - \frac{k r ^{2}}{8}

= - \frac{k r ^{2}}{8}

Add a constant to the solution

= - \frac{k r ^{2}}{8} + C

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What is the integral of k/r (-(r^2)/4) ?
The integral of k/r (-(r^2)/4) is -(kr^2)/8+C