`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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`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`