Question #0b19a

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Answer 1 · 2016-06-06T07:36:05

$int ((2x+13)dx)/(x^2+6x+13) =$
$ln (x^2+6x+13)+7/2*arctan ((x+3)/2)+C$

From the given $int ((2x+13)dx)/(x^2+6x+13)$

We can expand the integrand this way

$int ((2x+13)dx)/(x^2+6x+13)=int((2x+6+7)dx)/(x^2+6x+13)$

Then we split the fraction

$int ((2x+13)dx)/(x^2+6x+13)=int((2x+6)dx)/(x^2+6x+13) +int((7)dx)/(x^2+6x+13)$

We also know that $x^2+6x+13=(x+3)^2+4$
So that we can also write the integrals this way

$int ((2x+13)dx)/(x^2+6x+13)=int((2x+6)dx)/(x^2+6x+13) +7*int(dx)/((x+3)^2+4)$

Use now use the formulas

$int (du)/u=ln u$ and $int (du)/(u^2+a^2)=1/a*arctan(u/a)$

it follows

$int ((2x+13)dx)/(x^2+6x+13)=$

$ln (x^2+6x+13) +7/2*arctan ((x+3)/2)+C$

God bless.....I hope the explanation is useful.