integral help

**jzellt** · December 8th 2009, 07:41 PM

Evalute the integral:

((1-x)/x)^2 dx

I think I need to do integration by parts by I don't know the steps... please help

Thanks.

**Prove It** · December 8th 2009, 07:47 PM

Originally Posted by jzellt

Evalute the integral:

((1-x)/x)^2 dx

I think I need to do integration by parts by I don't know the steps... please help

Thanks.

No need for integration by parts.

Recall that $\frac{1 - x}{x} = \frac{1}{x} - 1 = x^{-1} - 1$ .

So $\left(\frac{1 - x}{x}\right)^2 = (x^{-1} - 1)^2$

$= x^{-2} - 2x^{-1} + 1$ .

Therefore

$\int{\left(\frac{1 - x}{x}\right)^2\,dx} = \int{x^{-2} - 2x^{-1} + 1\,dx}$

$= -x^{-1} - 2\ln{|x|} + x + C$

$= x - \frac{1}{x} - 2\ln{x} + C$ .

**pickslides** · December 8th 2009, 07:48 PM

$<br /> \int \left(\frac{1-x}{x}\right)^2~dx = \int \left(\frac{1}{x}-1\right)^2~dx= \int \left(\frac{1}{x^2}-\frac{2}{x}+1\right)~dx<br />$

now integrate it term by term.

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