integral x^3/(x^2-2) dx

I did u sub and got (x^2-2)/2 + ln(x^2-2) + C but wolfram alpha says it's x^2/2 + ln(x^2-2) + C

did I do something wrong? does u sub here don't work?

Question

integral x^3/(x^2-2) dx

I did u sub and got (x^2-2)/2 + ln(x^2-2) + C but wolfram alpha says it's x^2/2 + ln(x^2-2) + C

did I do something wrong? does u sub here don't work?

zarkon · Answer

those answers are the same

anonymous · Answer

how? I can see you could separate the fraction for the second one but where does the constant -2/2 go?

zarkon · Answer

into the other constant

anonymous · Answer

wouldn't that make the two constant different? +C and +C-1?

zarkon · Answer

$$(x^2-2)/2 +\ln(x^2-2) + C$$
$$=x^2/2 +\ln(x^2-2) + C-1$$
$$=x^2/2 +\ln(x^2-2) + C_2$$

anonymous · Answer

oh so because we don't know the value of original constant, we can assume any changes in the constant would not change the function?

zarkon · Answer

It is not that we don't know the original constant...the antiderivative of a function $f$ is a Function $F$ such that $F'=f$

both functions above satisfy this.

anonymous · Answer

oh I think I see now
thanks :)