What is the integral of 8/(1+9r^2) ?

The integral of 8/(1+9r^2) is 8/3 arctan(3r)+C

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integral of 8/(1+9r^2)

\int \frac{8}{1 + 9 r ^{2}} d r

\frac{8}{3} arctan (3 r) + C

Solution steps

\int \frac{8}{1 + 9 r ^{2}} d r

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 + 9 r ^{2}} d r

Apply Integral Substitution

= 8 \cdot \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 8 \cdot \frac{1}{3} arctan (u)

Substitute back

u = 3 r

= 8 \cdot \frac{1}{3} arctan (3 r)

Simplify

8 \cdot \frac{1}{3} arctan (3 r) : \frac{8}{3} arctan (3 r)

= \frac{8}{3} arctan (3 r)

Add a constant to the solution

= \frac{8}{3} arctan (3 r) + C

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What is the integral of 8/(1+9r^2) ?
The integral of 8/(1+9r^2) is 8/3 arctan(3r)+C