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Misc integral

  • Thread starter nameVoid
  • Start date
  • #1
241
0
[tex]
\int \frac{ln(x+1)}{x^2}dx
[/tex]
[tex]
u=x+1
[/tex]
[tex]
\int\frac{lnu}{(u-1)^2}du
[/tex]
[tex]
\int \frac{A}{u-1}+\frac{B}{(u-1)^2}du
[/tex]
[tex]
lnu=A(u-1)+B
[/tex]
[tex]
B=0,A=ln2
[/tex]
[tex]
\int \frac{ln2}{u-1}du
[/tex]
[tex]
ln2ln|x|+C
[/tex]
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
You have to do integration by parts.
Int[ln(x+1)/x^2] = ln(x+1)*intg(1/x^2) - Intg{[Intg(1/x^2)*d/dx[ln(x+1)]}
Now proceed.
 
  • #3
241
0
where is the error
 
  • #4
rl.bhat
Homework Helper
4,433
7
I = -1/x*ln(x+1) - int[(-1/x)*1/(x+1)]
= -1/x*ln(x+1) + int(1/x) - int1/(x+1)
Now find the integration.
In your partial factor, there is no ln function in the right hand side. So you can't equate the coefficients.
 

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