Question

How do you integrate `ln(x^2 + 13x + 40) dx`?

Answer 1

`(2x+13)/2*Ln(x^2+13x+40)-2x+3/2Ln((x+8)/(x+5))+C`

Explanation:

`int Ln(x^2+13x+40)*dx`

=`xLn(x^2+13x+40)-int x*((2x+13)*dx)/(x^2+13x+40)`

=`xLn(x^2+13x+40)-int ((2x^2+13x)*dx)/(x^2+13x+40)`

=`xLn(x^2+13x+40)-int ((2x^2+26x+80-13x-80)*dx)/(x^2+13x+40)`

=`xLn(x^2+13x+40)-int 2dx+int ((13x+80)*dx)/(x^2+13x+40)`

=`xLn(x^2+13x+40)-2x+C1+int ((52x+320)*dx)/(4x^2+52x+160)`

=`xLn(x^2+13x+40)-2x+C1+int ((26x+160)*2dx)/((2x+13)^2-3^2)`

`A=int ((26x+160)*2dx)/((2x+13)^2-3^2)`

After using `2x+13=3secy`, `2dx=3secy*tany*dy` and `x=(3secy-13)/2` transforms, `A` became

`A=int ((26*(3secy-13)/2+160)*3secy*tany*dy)/(9(tany)^2)`

=`int ((39secy-9)*secy*dy)/(3tany)`

=`13int((secy)^2*dy)/(tany)-3int (secy*dy)/tany`

=`13Ln(tany)-3int cscy*dy`

=`13/2Ln((tany)^2)-3int (cscy*(cscy+coty)*dy)/(cscy+coty)`

=`13/2Ln((secy)^2-1)+3Ln(cscy+coty)`

=`13/2Ln((secy)^2-1)+3Ln((secy+1)/tany)`

=`13/2Ln((secy)^2-1)+3Ln((secy+1)/sqrt((secy)^2-1))`

=`13/2Ln((secy)^2-1)+3/2Ln((secy+1)^2/((secy)^2-1))`

=`13/2Ln((secy)^2-1)+3/2Ln((secy+1)/(secy-1))`

After using `2x+13=3secy` and `secy=(2x+13)/3` inverse transforms, I found

`A=13/2Ln((2x+13)/3)^2-1)+3/2Ln(((2x+13)/3+1)/((2x+13)/3-1))`

=`13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))`

Thus,

`int Ln(x^2+13x+40)*dx`

=`xLn(x^2+13x+40)-2x+C1+13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))`

=`xLn(x^2+13x+40)-2x+13/2Ln((x^2+13x+40)`+`3/2Ln((x+8)/(x+5))+C`
=`(2x+13)/2*Ln(x^2+13x+40)-2x+3/2Ln((x+8)/(x+5))+C`

Note: `C=C1-13/2*Ln(4/9)=C1+13Ln(3/2)`