int4/(4x^2+4x+65)dx
Take the constant out,
=4int1/(4x^2+4x+65)dx
Complete the square for the denominator,
=4int1/((2x+1)^2+64)dx
Apply the integral substitution: u=(2x+1)
=>du=2dx
=>dx=(du)/2
=4int1/(u^2+8^2)((du)/2)
=2int1/(u^2+8^2)du
Now use the standard integral:int1/(x^2+a^2)dx=1/aarctan(x/a)
=2(1/8)arctan(u/8)
=1/4arctan(u/8)
Substitute back u=(2x+1) and add a constant C to the solution,
=1/4arctan((2x+1)/8)+C
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