C3/C4 Integrating Fractions help

Hi
just doing a c4 paper. A part of the question requires me to integrate 1/(1-x)
Surely this would be ln(1-x). But the answer is in fact -ln(1-x)

i know that theres a rule where the integral of f'(X)/f(x)= ln(fx). how do i 'play around' with this rule to get the correct answer?

thanks

Reply 1

Middriver

6

Original post by starwarsjedi123

Hi
just doing a c4 paper. A part of the question requires me to integrate 1/(1-x)
Surely this would be ln(1-x). But the answer is in fact -ln(1-x)

i know that theres a rule where the integral of f'(X)/f(x)= ln(fx). how do i 'play around' with this rule to get the correct answer?

thanks

Dy/dx of 1-X =-1

Reply 2

starwarsjedi123

OP

4

so say the f'x/f(x) is almost true, i just put the derivative of the denominator,i.e. f(x), in front of the ln(fx)??????

Reply 3

Middriver

6

Original post by starwarsjedi123

so say the f'x/f(x) is almost true, i just put the derivative of the denominator,i.e. f(x), in front of the ln(fx)??????

You put whatever makes the numerator - the derivative of the denominator infront of Lnx.

Reply 4

Zacken

22

Original post by starwarsjedi123

Hi
just doing a c4 paper. A part of the question requires me to integrate 1/(1-x)
Surely this would be ln(1-x). But the answer is in fact -ln(1-x)

i know that theres a rule where the integral of f'(X)/f(x)= ln(fx). how do i 'play around' with this rule to get the correct answer?

thanks

This post I've written up might help:

It's also quite nice (not sure if this is C4 or not, apologies if it isn't), let's say I want to integrate this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int xe^{6x^2} \, \mathrm{d}x\end{equation*}

I know that it looks very similar to a

f'(x)e^{f(x)}

form which integrates to

e^{f(x)}

(you know why, right?) but it's not quite. Instead we have

\frac{\mathrm{d}}{\mathrm{d}x} (6x^2) = 12x

, so what I really want is a

12xe^{6x^2}

, but that's not what I have. Instead of crying over that and being sad, let's pop in the

12

there because we're badass (but also fix our damage, because we're nice like that!), what I mean is:

Unparseable latex formula:

\displaystyle[br] \begin{equation*}\int xe^{6x^2} \, \mathrm{d}x = \frac{1}{12} \int 12xe^{6x^2} \, \mathrm{d}x = \frac{1}{12}e^{6x^2 } + c\end{equation*}

If you want to have a go, have a stab at

\int \frac{1}{1 + 5x} \, \mathrm{d}x

. We know that it's almost of the form

\frac{f'(x)}{f(x)}

which integrates nicely to

\ln f(x)

, but it's not quite in that form, what do you need to do to get it in that form?

So basically, look at the derivative of the denominator, manipulate the numerator to make it the derivative of the numerator but make sure that you adjust it to keep it the same.

So in your case, you want the numerator to be -1. Shove in a -1 in there but add in another -1 to counter it.

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int \frac{1}{1-x} \, \mathrm{d}x = -\int\frac{-1}{1-x} \, \mathrm{d}x \end{equation*}

Now the numerator is the perfect derivative of the denominator so you can go ahead and do:

\int \frac{\,\mathrm{d}x}{1-x} = -\ln|1-x| + \mathcal{C}

C3/C4 Integrating Fractions help

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