Integrate Partial Fractions: x^3/(x^2-1)

AI Thread Summary
To integrate x^3/(x^2 - 1) using partial fractions, the numerator's degree exceeds that of the denominator, so polynomial long division is necessary. Dividing gives the result x + 1/(x-1) - 1/((x+1)(x-1)). For the definite integral from 1 to e of (ln(x))/x^2, a suitable substitution is t = ln(x), leading to dt = (1/x)dx. This substitution simplifies the integral but may still require integration by parts afterward. The discussion emphasizes the importance of recognizing when to apply polynomial long division before using partial fractions.
Firepanda
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Integrate using partial fractions:

(int) (x^3)/(x^2 -1) dx

I have put into the form (int) (x^3)/((x-1)(x+1)) dx

I thought partial fractions had this property:

'Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator.'

And this obviously doesn't, so how do you do it?
 
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Firepanda said:
Integrate using partial fractions:

(int) (x^3)/(x^2 -1) dx

I have put into the form (int) (x^3)/((x-1)(x+1)) dx

I thought partial fractions had this property:

'Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator.'

And this obviously doesn't, so how do you do it?

Divide! What is x3 divided by x2- 1? What are the quotient and remainder?
 
ahhh

so it =

x + 1/(x-1) - 1/(x+1)(x-1)

:D
 
While I am here I have another question:

(int) 1 to e (ln(x))/x^2 dx

THe question says, use a suitable substitution to evaluate the definite integral, I can do it by parts but I don't want to lose marks for it.

Can any1 suggest the substitution to get me started?
Thx
 
Firepanda said:
While I am here I have another question:

(int) 1 to e (ln(x))/x^2 dx

THe question says, use a suitable substitution to evaluate the definite integral, I can do it by parts but I don't want to lose marks for it.

Can any1 suggest the substitution to get me started?
Thx
This should be in the Calculus section.

Verifying your problem ...

\int_1^e\frac{\ln x}{x^2}dx
 
rocophysics said:
This should be in the Calculus section.

Verifying your problem ...

\int_1^e\frac{\ln x}{x^2}dx

Yep that's the one
 
Firepanda said:
Yep that's the one
Substitutions ...

t=\ln x \rightarrow e^t=x

dt=\frac 1 x dx

You will need to do parts afterwards.
 

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