# Problem with an Integral

1. Apr 21, 2005

### Pyrrhus

$$\int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx$$

Here's what i have done

$$\int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx + \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx$$

so

$$I_{1} = I_{2} + I_{3}$$

where

$$I_{1} = \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx$$

$$I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx$$

$$I_{3} = \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx$$

Now,

$$I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx$$

multiplying by $\frac{\ln{y}}{\ln{y}}$

$$I_{2} = \frac{\ln{y}}{\ln{y}} \int \frac{1}{\frac{1}{y} + x \ln {y}} dx$$

$$I_{2} = \frac{1}{\ln{y}} \int \frac{\ln{y}}{\frac{1}{y} + x \ln {y}} dx$$

Using substitution $u = \frac{1}{y} + x \ln {y}$ and therefore $du = \ln{y} dx$

Now,

$$I_{2} = \frac{1}{\ln{y}} \int \frac{1}{u} du$$

$$I_{2} = \frac{1}{\ln{y}} \ln{|\frac{1}{y} + x \ln {y}|}$$

Any ideas about $I_{3}$ or the whole integral?

2. Apr 21, 2005

### whozum

I dont think thats fair game.

3. Apr 21, 2005

### dextercioby

There you go

$$\int \frac{1-xe^{xy}}{\frac {1}{y}+x\ln y}dx=\allowbreak \frac{\ln \left( \frac {1}{y}+x\ln y\right) }{\ln y}-\frac {1}{y\ln y}e^{xy}-\frac {1}{y\ln^{2}y}e^{-\frac 1{\ln y}}\func{Ei}\left( 1,-xy-\frac 1{\ln y}\right) +C$$

Daniel.

4. Apr 22, 2005

### Pyrrhus

Thanks Dexter , i guess my first integral was correct ($I_{2}$) but the second one ($I_{3}$) , whoa what a mess...

By the way, any idea for solving the $I_{3}$ ?, i haven't been able to see anything.... maybe i can rewrite it someway... i'll let you all know if i can.

Whozum? what do you mean? y is being kept as a constant...

Last edited: Apr 22, 2005
5. Apr 22, 2005

### dextercioby

You'll have to dig it somehow (make algebraic calculations and prossibly changes of integration variable) as to get it in the standard form of the Ei function.

Daniel.