You should write the fraction `(4x + 1)/(2x^2 + 4x + 10) = (4x)/(2x^2 + 4x + 10) + 1/(2x^2 + 4x + 10).`

Integrating the fraction yields:

`int ((4x + 1)dx)/(2x^2 + 4x + 10) = int (4x dx)/(2x^2 + 4x + 10) + int dx/(2x^2 + 4x + 10)`

You should notice that differentiating the denominator of the fraction yields:

`(2x^2 + 4x + 10)'= 4x + 4`

Hence, adding and subtracting 4 to numerator of the first fraction yields:

`int ((4x + 4 - 4)dx)/(2x^2 + 4x + 10) = int ((4x + 4)dx)/(2x^2 + 4x + 10) -4 int (dx)/(2x^2 + 4x + 10)`

`int ((4x + 1)dx)/(2x^2 + 4x + 10) = int ((4x + 4)dx)/(2x^2 + 4x + 10) -4 int (dx)/(2x^2 + 4x + 10) +int dx/(2x^2 + 4x + 10)`

`int ((4x + 1)dx)/(2x^2 + 4x + 10) = int ((4x + 4)dx)/(2x^2 + 4x + 10) -3int (dx)/(2x^2 + 4x + 10)`

You should use substitution to solve the first integral such that:

`2x^2 + 4x + 10 = y`

Differentiating yields: `(4x+4)dx = dy`

`int ((4x + 1)dx)/(2x^2 + 4x + 10) = int (dy)/(y) -3int (dx)/(2x^2 + 4x + 10)`

You should solve the second integral completing the square at denominator such that:

`3int (dx)/(2((x^2 + 2x + 1) - 1 + 5))) = (3/2)*int (dx)/((x+1)^2 +2^2))`

You need to use the formula

`int dx/(x^2 + a^2) = (1/a)*arctan (x/a) + c`

`(3/2)*int (dx)/((x+1)^2+ 2^2)) = (3/4)*arctan ((x+1)/2) + c`

**Evaluating the given integral yields: `int ((4x + 1)dx)/(2x^2 + 4x + 10) = ln(2x^2 + 4x + 10)-(3/4)*arctan ((x+1)/2) + c` .**

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