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

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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` .**

(4x + 1)/(2x^2 + 4x + 10)

Notice that the derivative of the denominator is 4x + 4.

=(4x + 4)/(2x^2 + 4x + 10) - 3/(2x^2 + 4x + 10)

We can integrate the first term with substitution now.

u=2x^2 + 4x + 10, du = 4x + 4

=integral(1/u) du - 3 * integral(1/(2x^2 + 4x + 10)) dx

=ln(u) - 3 * integral(1/(2x^2 + 4x + 10)) dx

Complete the square on the bottom right and substitute to get a sum of squares

=ln(u) - 3 * integral(1/((sqrt(2)x +sqrt(2))^2 + 8)) dx

s = sqrt(2)x + sqrt(2), ds = sqrt(2)

=ln(u) - 3/sqrt(2) * integral(1/s^2 + 8) ds

Use the table of integrals formula (pretty much every book has this formula) for 1/(x^2 + u^2)

=ln(u) - 3/sqrt(2) * arctan(s/sqrt(8))/sqrt(8) + C

Simplify

=ln(u) - 3/4 * arctan(s/sqrt(8)) + C

Substitute

=ln(2(x^2 + 2x + 5)) - 3/4 * arctan((sqrt(2)x + sqrt(2))/sqrt(8)) + C

Simplify again

=ln(x^2 + 2x + 5) - 3/4 * arctan(x/2 + 1/2) + C

Note that the ln(2) is changed from multiplication inside the logarithm to a sum outside the logarithm (ln(xy) = ln(x) + ln(y)). Since ln(2) is a constant, it is absorbed by C.