What is the integral of (15)/(1+16r^2) ?

The integral of (15)/(1+16r^2) is 15/4 arctan(4r)+C

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integral of (15)/(1+16r^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

u = 4 r

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (15)/(1+16r^2) ?
The integral of (15)/(1+16r^2) is 15/4 arctan(4r)+C