U-Substitution

[dsheller]

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!

[Glaysher]

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

[Mattywoo2]

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)

[Dross]

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.

[Glaysher]

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.

Yes, how annoying