Question

Question #0b19a

Answer 1

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

Explanation:

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.