What is the integral of 1/(25x^2+36) ?

The integral of 1/(25x^2+36) is 1/30 arctan(5/6 x)+C

integral of 1/(25x^2+36)

\int \frac{1}{25 x ^{2} + 36} d x

\frac{1}{30} arctan (\frac{5}{6} x) + C

Solution steps

\int \frac{1}{25 x ^{2} + 36} d x

Apply Integral Substitution

= \int \frac{1}{30 ( 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}{30} \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}{30} arctan (u)

Substitute back

u = \frac{5}{6} x

= \frac{1}{30} arctan (\frac{5}{6} x)

Add a constant to the solution

= \frac{1}{30} arctan (\frac{5}{6} x) + C

\frac{d}{d x} (m x^{m - 1})

\frac{\partial}{\partial x} (\frac{( x - 3 y )}{x + 3 y})

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

a re a y = x, 0, 1

d er i v a t i v e f (t) = e^{7 t \cdot s i n (2 t)}

What is the integral of 1/(25x^2+36) ?
The integral of 1/(25x^2+36) is 1/30 arctan(5/6 x)+C