`int 1 /(25+4x^2) dx` Find the indefinite integral

`int1/(25+4x^2)dx`

Let's transform the denominator of the integral,

`int1/(25+4x^2)dx=int1/(4(x^2+25/4))dx`

Take the constant out,

`=1/4int1/(x^2+(5/2)^2)dx`

Now use the standard integral:`int1/(x^2+a^2)dx=1/aarctan(x/a)`

`=1/4(1/(5/2))arctan(x/(5/2))`

simplify and add a constant C to the solution,

`=(1/4)(2/5)arctan((2x)/5)+C`

`=1/10arctan((2x)/5)+C`

`int1/(25+4x^2)dx`

Let's transform the denominator of the integral,

`int1/(25+4x^2)dx=int1/(4(x^2+25/4))dx`

Take the constant out,

`=1/4int1/(x^2+(5/2)^2)dx`

Now use the standard integral:`int1/(x^2+a^2)dx=1/aarctan(x/a)`

`=1/4(1/(5/2))arctan(x/(5/2))`

simplify and add a constant C to the solution,

`=(1/4)(2/5)arctan((2x)/5)+C`

`=1/10arctan((2x)/5)+C`