[tex] \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex](adsbygoogle = window.adsbygoogle || []).push({});

Here's what i have done

[tex] \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx + \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

so

[tex] I_{1} = I_{2} + I_{3} [/tex]

where

[tex] I_{1} = \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

[tex] I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

[tex] I_{3} = \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

Now,

[tex] I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

multiplying by [itex] \frac{\ln{y}}{\ln{y}} [/itex]

[tex] I_{2} = \frac{\ln{y}}{\ln{y}} \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

[tex] I_{2} = \frac{1}{\ln{y}} \int \frac{\ln{y}}{\frac{1}{y} + x \ln {y}} dx [/tex]

Using substitution [itex] u = \frac{1}{y} + x \ln {y} [/itex] and therefore [itex] du = \ln{y} dx [/itex]

Now,

[tex] I_{2} = \frac{1}{\ln{y}} \int \frac{1}{u} du [/tex]

[tex] I_{2} = \frac{1}{\ln{y}} \ln{|\frac{1}{y} + x \ln {y}|} [/tex]

Any ideas about [itex] I_{3} [/itex] or the whole integral?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Problem with an Integral

**Physics Forums | Science Articles, Homework Help, Discussion**