# [SOLVED] Integration of Inverse of f(x)

by JayKo
Tags: integration, inverse, solved
 P: 128 $$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ 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
 HW Helper P: 858 yes, factorise and then partial fractions is ok
 P: 128 wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks
HW Helper
P: 858

## [SOLVED] Integration of Inverse of f(x)

firstly you can further factorise y^2-6y+5
 P: 128 $$\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ =$$\frac{1}{y^{2}(y^{2}-6y+5)}$$ =$$\frac{1}{y^{2}(y-1)(y-5)}$$ 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
P: 128
 Quote by mjsd firstly you can further factorise y^2-6y+5
i know, is just that i m slow using teX, please gimme a min.thanks
 HW Helper P: 858 I can tell you the answer.... no need use latex for such simple expression most likely you have treated the y^2 bit incorrectly
 P: 128 after i substitute A,B,C,and D. i can get back my original 1/f(x)
 HW Helper P: 858 looks ok so far
P: 128
 Quote by 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
lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.
 HW Helper P: 858 FYI i've got A=6/25 B=1/5 C=-1/4 D=1/100
P: 128
 Quote by mjsd looks ok so far
if so, then
$$\frac{1}{y^{2}(y-1)(y-5)}$$=$$\frac{6}{25y}$$+$$\frac{1}{5y^{2}}$$+$$\frac{3}{10(y-1)}$$-$$\frac{87}{50(y-5}$$
P: 128
 Quote by mjsd FYI i've got A=6/25 B=1/5 C=-1/4 D=1/100
OOops,gimme a minute,i recheck :)
P: 128
 Quote by mjsd FYI i've got A=6/25 B=1/5 C=-1/4 D=1/100
you are right! :D :D :D thanks a lot.for pointing it out !:D
 P: 128 $$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ 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}}$$ dy = 6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y) then i m stuck to make y the subject. anyone to point out the mistake.thanks
 P: 128 anyone, can show me the right way?
 P: 128 another differential eq. question. y'=1+xy find y. how to go about it. i m clueless my method of multiply dx to (1+xy) doesn't work in this case.
 P: 414 You can solve this by the solution for Linear Differential equation. The equation you have is: $$\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..

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