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Integration question

Hey, I'd be grateful for any help on integrating this

(x^2 + 1) / (x^3 + 3x +2)

I see that the derivative of (x^3 + 3x +2) is 3(x^2 +1) which I feel should make it easier but I can't see the trick?
Original post by sqrt123
Hey, I'd be grateful for any help on integrating this

(x^2 + 1) / (x^3 + 3x +2)

I see that the derivative of (x^3 + 3x +2) is 3(x^2 +1) which I feel should make it easier but I can't see the trick?


Rewrite

∫x2+1x3+3x+2.dx\displaystyle \int \dfrac{x^2+1}{x^3+3x+2}.dx

as

13∫3(x2+1)x3+3x+2.dx\displaystyle \dfrac{1}{3} \int \dfrac{3(x^2+1)}{x^3+3x+2}.dx.
Reply 2
Avatar for sqrt123
sqrt123
OP
10
Original post by RDKGames
Rewrite

∫x2+1x3+3x+2.dx\displaystyle \int \dfrac{x^2+1}{x^3+3x+2}.dx

as

13∫3(x2+1)x3+3x+2.dx\displaystyle \dfrac{1}{3} \int \dfrac{3(x^2+1)}{x^3+3x+2}.dx.

Thank you :smile:
I've got that far but I don't know how to carry on... I've tried by parts and substitution of u = (x^2 +1) :s-smilie:
Original post by sqrt123
Thank you :smile:
I've got that far but I don't know how to carry on... I've tried by parts and substitution of u = (x^2 +1) :s-smilie:


But you literally just told me that the derivative of x3+3x+2x^3+3x+2 is 3(x2+1)3(x^2+1) so that second form is an integral of the form f'(x)/f(x). Doesn't it sound familiar?

If you really want to use a sub here then just use u=x3+3x+2u=x^3+3x+2.
Reply 4
Avatar for sqrt123
sqrt123
OP
10
Original post by RDKGames
But you literally just told me that the derivative of x3+3x+2x^3+3x+2 is 3(x2+1)3(x^2+1) so that second form is an integral of the form f'(x)/f(x). Doesn't it sound familiar?

If you really want to use a sub here then just use u=x3+3x+2u=x^3+3x+2.

Ohhh of course! I completely missed that :colondollar:
I get 1/3 ln(x^3 + 3x +2) + c, thanks for your help :smile:
Original post by sqrt123
Ohhh of course! I completely missed that :colondollar:
I get 1/3 ln(x^3 + 3x +2) + c, thanks for your help :smile:


Yep, modulus brackets around the cubic are a must :smile:

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