# Integration of Inverse of f(x)

1. Jan 13, 2008

### JayKo

[SOLVED] Integration of Inverse of f(x)

$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

$$\frac{1}{4y^{3}-18y^{2}+10y}$$(ln|$${y^{4}-6y^{3}+5y^{2}}$$|)

is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks

2. Jan 13, 2008

### mjsd

yes, factorise and then partial fractions is ok

3. Jan 13, 2008

### JayKo

wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks

4. Jan 13, 2008

### mjsd

firstly you can further factorise y^2-6y+5

5. Jan 13, 2008

### JayKo

$$\frac{1}{y^{4}-6y^{3}+5y^{2}}$$

=$$\frac{1}{y^{2}(y^{2}-6y+5)}$$

=$$\frac{1}{y^{2}(y-1)(y-5)}$$

So, by Partial fraction, is this correct?

$$\frac{1}{y^{2}(y-1)(y-5)}$$
=$$\frac{A}{y}$$+$$\frac{B}{y^{2}}$$+$$\frac{C}{y-1}$$+$$\frac{D}{y-5}$$

and then, by comparing coefficient of y^3, y^2, y.

(A+C+D){y^{3}}=0
(-6A+B-5C-D){y^{2}}=0
(5A-6B)y=0
answer is A=6/25, B=1/5, C=3/10. D=-87/50.

but doubt my answer. anyone can point out?thanks

Last edited: Jan 13, 2008
6. Jan 13, 2008

### JayKo

i know, is just that i m slow using teX, please gimme a min.thanks

7. Jan 13, 2008

### mjsd

I can tell you the answer....

no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly

Last edited: Jan 13, 2008
8. Jan 13, 2008

### JayKo

after i substitute A,B,C,and D.

i can get back my original 1/f(x)

9. Jan 13, 2008

### mjsd

looks ok so far

10. Jan 13, 2008

### JayKo

lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.

11. Jan 13, 2008

### mjsd

FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100

12. Jan 13, 2008

### JayKo

if so, then
$$\frac{1}{y^{2}(y-1)(y-5)}$$=$$\frac{6}{25y}$$+$$\frac{1}{5y^{2}}$$+$$\frac{3}{10(y-1)}$$-$$\frac{87}{50(y-5}$$

Last edited: Jan 13, 2008
13. Jan 13, 2008

### JayKo

OOops,gimme a minute,i recheck :)

14. Jan 13, 2008

### JayKo

you are right! :D :D :D thanks a lot.for pointing it out !:D

15. Jan 13, 2008

### JayKo

$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right?

but this problem is actually make y the subject in this equation, how to go about it further?
$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy = $$\int\$$ dt

the original question is $$\frac{dy}{dt}$$=$$y^{4}-6y^{3}+5y^{2}$$ find y(t).

if $$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y)
then i m stuck to make y the subject. anyone to point out the mistake.thanks

Last edited: Jan 13, 2008
16. Jan 13, 2008

### JayKo

anyone, can show me the right way?

17. Jan 13, 2008

### JayKo

another differential eq. question.

y'=1+xy find y. how to go about it. i m clueless
my method of multiply dx to (1+xy) doesn't work in this case.

18. Jan 13, 2008

### rohanprabhu

You can solve this by the solution for Linear Differential equation. The equation you have is:

$$\frac{dy}{dx} = 1 + xy$$

which can be written as:

$$\frac{dy}{dx} + y(-x) = 1$$

which is similar to:

$$\frac{dy}{dx} + yP = Q$$

Here, $P = (-x); Q = 1$

So, you have:

$$I.F = e^{\int (-x)dx}$$
$$I.F = e^{-\frac{x^2}{2}}$$

And hence, your solution is given by:

$$ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C$$

But, it's gonna be a real pain with this:

$$\int e^{\frac{-x^2}{2}}dx$$

a method to which i can't think of right now. Maybe there's some other way..

Last edited: Jan 13, 2008
19. Jan 13, 2008

### Gib Z

That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.

20. Jan 14, 2008

### JayKo

great, this is something new to me, just started my linear algebra course.thanks.will study your solution