1. Integration by substitution

I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

I can't seem to bash it out though.

u=1/x

du/dx=-1/x^2

when I try to substitute it all goes to pot.

I need the method before moving onto the real work!

Thanks

TF

2. Originally Posted by Taylor Fylde
I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

I can't seem to bash it out though.

u=1/x

du/dx=-1/x^2

when I try to substitute it all goes to pot.

I need the method before moving onto the real work!

Thanks

TF
It's easier to do it using partial fractions, but if you insist:

$\displaystyle du = -\frac{1}{x^2}~dx$

$\displaystyle dx = -x^2~du$

$\displaystyle dx = -\frac{1}{u^2}~du$

So
$\displaystyle \int \frac{1}{x(1 - ax)}~dx = -\int \frac{u}{u^2 \left ( 1 - \frac{a}{u} \right ) }~du$

$\displaystyle = -\int \frac{1}{u - a}~du$

$\displaystyle = -ln \left | u - a \right | + C$

$\displaystyle = -ln \left | \frac{1}{x} - a \right | + C$

-Dan

3. Hello,

Originally Posted by Taylor Fylde
I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

I can't seem to bash it out though.

u=1/x

du/dx=-1/x^2

when I try to substitute it all goes to pot.

I need the method before moving onto the real work!

Thanks

TF
Ok for the substitution... Let's study it :

$\displaystyle \boxed{\frac{1}{x(1-ax)}}$

As you said : $\displaystyle \frac{du}{dx}=-\frac{1}{x^2}$

But we want it with respect to u.

$\displaystyle \frac{du}{dx}=-\frac{1}{x^2}=-\left(\frac 1x\right)^2=-u^2$

$\displaystyle \implies dx=-\frac{du}{u^2}$

Hence, the substitution yields :

\displaystyle \begin{aligned} \int \frac{1}{x(1-ax)} dx &= \int \frac 1x \cdot \frac{1}{1-ax} dx \\ &= \int u \cdot \frac{1}{1-\frac au} \cdot \left(-\frac{1}{u^2}\right) du \\ &= -\int \frac 1u \cdot \frac{1}{1-\frac au} du \\ &= -\int \frac{1}{u(1-\frac au)} du \end{aligned}
\displaystyle \begin{aligned} &= -\int \frac{1}{u-a} du \\ &= -\ln \left|u-a\right| + c \\ &= -\ln \left|\frac 1x-a\right| + c \end{aligned}

4. Many thanks (both of you).

I'm getting it, but I seem to have to substitute a bit at a time. Each time I write it out, I substitute in a different order. Time for bed - let it ferment over night.

TF