# evaluate the following: integral of (4x + 1)/(2x^2 + 4x + 10)dxfind the integral of (4x + 1)divide by (2x^2 + 4x + 10)

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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