what is the general antiderivative of ∫ (8x) / (x^2 + 1) dx?

Answer 1

I = 4 ∫ 2 x / (x ² + 1) dx
I = 4 log (x ² + 1) + C

Answer 2

The general antiderivative is 4 ln (x^2 + 1) + const.

That's because, by the chain rule,

d/dx [4 ln (x^2 + 1) + const.] = 4 * 2x / (x^2 + 1) = 8x / (x^2 + 1).

[Modulus signs for the ln term are redundant since (x^2 + 1) is always positive, of course.]

Live long and prosper.

Answer 3

∫ (8x) / (x^2 + 1) dx

Use substitution

u = (x^2 + 1)
du = 2x...so,

4∫ (2x) / (x^2 + 1) dx
4∫ 1 / u du
= 4lnlul+c
= 4lnlx^2 + 1l+c...--->General Solution.