Alright I have the integral x^3 / 1 + x^2
Thus far I have gotten to
1/2 integral u / 1 + u du but after that I have no idea what to do, or if that is even right. If anyone could shed some light on how to do this that would be great!
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Alright I have the integral x^3 / 1 + x^2
Thus far I have gotten to
1/2 integral u / 1 + u du but after that I have no idea what to do, or if that is even right. If anyone could shed some light on how to do this that would be great!
I assume you mean integral of x3/(1 + x2) dx
Try substitution u = 1 + x2
Then du/dx = 2x
So dx = (1/2x) du
So integral becomes integral of x3/(u2x) du
integral of (x2/u) du
But u = 1 + x2
So x2 = u - 1
so integral of [(u - 1)/u] du
integral of 1 - (1/u) du
u - ln u + c
1 + x2 - ln (1 + x2) + c
I got the same answer by simply playing with the x's.
If you write the numerator x^3 as x(1+x²) - x then you have the form
x(1+x²)/(1+x²) - x/(1+x²)
x - x/(1+x²)
Integrating gives
1/2 x² - 1/2 ln(1+x²) + c
(our c's are different, that is all)
On the contrary, your answers are different because Glaysher left out a factor of 1/2 between the 5th and 6th line of his post. The answer is indeed 1/2x2 - 1/2ln(1+x2) + constant.Quote:
Originally Posted by Mattywoo2
Yes, how annoyingQuote:
Originally Posted by Dross