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