- #1
Saladsamurai
- 3,020
- 7
I am either making the same mistake repeatedly, or I can't factor!
I did this [tex]\int\ln(2x-3)dx[/tex] first by parts and then using the formula [tex]\int\ln u du=u\ln u-u+C[/tex]
By parts I got:
[tex]u=\ln(2x-3)[/tex] dv=dx
so [tex]=x\ln(2x-3)-\int\frac{2x}{2x-3}dx[/tex] and by long division:
[tex]=x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]
[tex]=x\ln(2x-3)-x-\frac{3}{2}\ln(2x+3)+C[/tex]
but I can't get that to match the formula result of
[tex](2x-3)\ln(2x-3)-(2x+3)+C[/tex]
Is the by parts correct?
Casey
I did this [tex]\int\ln(2x-3)dx[/tex] first by parts and then using the formula [tex]\int\ln u du=u\ln u-u+C[/tex]
By parts I got:
[tex]u=\ln(2x-3)[/tex] dv=dx
so [tex]=x\ln(2x-3)-\int\frac{2x}{2x-3}dx[/tex] and by long division:
[tex]=x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx[/tex]
[tex]=x\ln(2x-3)-x-\frac{3}{2}\ln(2x+3)+C[/tex]
but I can't get that to match the formula result of
[tex](2x-3)\ln(2x-3)-(2x+3)+C[/tex]
Is the by parts correct?
Casey