What is the integral of (15)/(16+r^2) ?

The integral of (15)/(16+r^2) is 15/4 arctan(r/4)+C

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integral of (15)/(16+r^2)

\int \frac{15}{16 + r ^{2}} d r

\frac{15}{4} arctan (\frac{r}{4}) + C

Solution steps

\int \frac{15}{16 + r ^{2}} d r

Take the constant out:

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

= 15 \cdot \int \frac{1}{16 + r ^{2}} d r

Apply Integral Substitution

= 15 \cdot \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 15 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 15 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{r}{4}

= 15 \cdot \frac{1}{4} arctan (\frac{r}{4})

Simplify

15 \cdot \frac{1}{4} arctan (\frac{r}{4}) : \frac{15}{4} arctan (\frac{r}{4})

= \frac{15}{4} arctan (\frac{r}{4})

Add a constant to the solution

= \frac{15}{4} arctan (\frac{r}{4}) + C

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What is the integral of (15)/(16+r^2) ?
The integral of (15)/(16+r^2) is 15/4 arctan(r/4)+C