Question about this integral

  • Thread starter alba_ei
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  • #1
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Homework Statement



[tex]\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta[/tex]

The Attempt at a Solution



i took [tex]u = ae^\theta-b[/tex] so [tex]e^\theta = \frac{u + b}{a}[/tex] then i substituded back into the integral and iget this

[tex]\int \frac{u + b + b}{u} \, du[/tex]

[tex]\int du +\int \frac{2b}{u} \, du[/tex]

[tex]= u \du + 2b \ln u +C[/tex]

[tex]= u + 2b \ln u +C[/tex]

[tex]= ae^\theta-b + 2b\ln (ae^\theta-b) [/tex]

but the answer of the book is
[tex]\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta = 2\ln (ae^\theta-b) - \theta + C [/tex]
what did i do wrong?
 
Last edited:

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
You didn't subsitute properly. You have to change dtheta too.
 
  • #3
236
0
write d0 as what it should equal to du
for example if u=x^2
du=2*xdx
 

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