Question

Integral of ln(x^(2)+15x+44) dx

Answer 1

Separate out
Int 15x = (15/2)x^2
Int 44 = 44x

Answer 2

\[U=\ln (x^2+15x+44)\] \[dU=(2x+15)/(x^2+15x+44)dx\] \[dV=dx\] \[V=x\] so:
\[I=xln(x^2+15x+44)-\int\limits_{}^{}(2x^2+15x)/(x^2+15x+44)dx\] now (2x^2+15x) you can write as :2(x^2+15x+44)-(15x+88) so now you have to evaluate two separate integrals. The first one is:\[I _{1}=\int\limits_{}^{}2dx=2x\] and the second one is:\[I _{2}=\int\limits_{}^{}(15x+88)/(x^2+15x+44)dx=\int\limits_{}^{}(15x+88)/((x+4)(x+11))\]dx now using the formula: \[(15x+88)/((x+4)(x+1))=A/(x+4)+B/(x+1)\] you get:

Answer 3

Oh, I missed the parentheses of ln

Answer 4

It's much more complicated than I thought :(

Answer 5

15x+88=(A+B)x+A+4B which gives you these equalities: A+B=15 and A+4B=88...solving this system gives you A=-28/3 and B=73/3 so:\[I _{2}=\int\limits_{}^{}(-28/3)/(x+4)dx+\int\limits_{}^{}73/3/(x+1)=(-28/3)\ln|x+4|+(73/3)\ln|x+1|\]

Answer 6

finally you get the integral:\[I=I _{1}-I _{_{2}}=2x+(28/3)\ln|x+4|-(73/3)\ln|x+1|+c\]

Answer 7

Thanks, I'm speechless!