[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**

Dismiss Notice

Join Physics Forums Today!

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

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

# Problem with an Integral

Loading...

Similar Threads for Problem Integral |
---|

A Maximization problem using Euler Lagrange |

A Maximization Problem |

I A question about Vector Analysis problems |

B Integrating Problem |

I Problems with this integral |

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