What is the integral of 1/(25+3x^2) ?

The integral of 1/(25+3x^2) is 1/(5sqrt(3))arctan((sqrt(3))/5 x)+C

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integral of 1/(25+3x^2)

\int \frac{1}{25 + 3 x ^{2}} d x

\frac{1}{5 3} arctan (\frac{3}{5} x) + C

Solution steps

\int \frac{1}{25 + 3 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{5 3 ( 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}{5 3} \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}{5 3} arctan (u)

Substitute back

u = \frac{3}{5} x

= \frac{1}{5 3} arctan (\frac{3}{5} x)

Add a constant to the solution

= \frac{1}{5 3} arctan (\frac{3}{5} x) + C

\int \frac{1}{x ^{2} 16 x ^{2} - 9} d x

\frac{\partial}{\partial x} (1 + x)

\frac{d v}{d x} = - 2 v - 32

\frac{\partial}{\partial y} (cos (x) y)

\frac{\partial}{\partial x} (e^{x + 1})

What is the integral of 1/(25+3x^2) ?
The integral of 1/(25+3x^2) is 1/(5sqrt(3))arctan((sqrt(3))/5 x)+C