What is the integral of 1/(9+16r^2) ?

The integral of 1/(9+16r^2) is 1/12 arctan(4/3 r)+C

Popular >

integral of 1/(9+16r^2)

\int \frac{1}{9 + 16 r ^{2}} d r

\frac{1}{12} arctan (\frac{4}{3} r) + C

Solution steps

\int \frac{1}{9 + 16 r ^{2}} d r

Apply Integral Substitution

= \int \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{12} arctan (u)

Substitute back

u = \frac{4}{3} r

= \frac{1}{12} arctan (\frac{4}{3} r)

Add a constant to the solution

= \frac{1}{12} arctan (\frac{4}{3} r) + C

\int \frac{72}{x ^{2} - 81} d x

\int (3 a - 2 x)^{2} d a

\int (\frac{9 - x ^{2}}{x}) d x

\int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{t ^{2} + 3 t} d t

\int \frac{( 2 x ^{3} + 1 )}{x} d x

What is the integral of 1/(9+16r^2) ?
The integral of 1/(9+16r^2) is 1/12 arctan(4/3 r)+C