how do you integrate: (3x)/(x^2+1) I think it has something to do with arctan but I don't know about the 3x ?

Question

integration using arctan

Answer 1

I don't think arctan will come into this one. If you make the change of variables u = x^2 + 1, you will get a solution in terms of logarithms:

Answer is ln[(x^2 + 1)^(3/2)].

Answer 2

Good intuition about using arctanx. However, in this particular circumstance, arctanx is unnecessary. Whenever you have an integral int(a, b, f(x)), where f(x) is a rational function (one polynomial being divided by another, nonzero polynomial) AND the degree of the polynomial in the numerator is 1 less than the degree of the polynomial in the denominator, you should be able to find a substitution that will allow you to integrate. Typically, the substitution will simply be to set a new variable equal to the polynomial in the denominator.

For example, with (3x)/(x^2+1),
let u = x^2 + 1.
Then
du = 2x dx,
and hence, multiplying both sides of this equation by 3 / 2,
(3/2)du = 3x dx.
Thus we can substitute these values into the original integral (if you are computing a definite integral, be sure to calculate your new bounds in terms of u) as
int((3x)dx/(x^2+1)) = (3/2)int(du/u)
= (3/2)ln|u| + C, where C is the constant of integration.
Rewriting this in terms of x, we get
(3/2)ln|u| + C = ln(|x^2 + 1|^(3/2)) + C.
Hope this helps!

Answer 3

No, don't use arctan to integrate it. Use natural logs and substitution method to integrate it

i.e. let u = x^2 +1
therefore du = 2x.dx
0.5 du = x.dx

Take out a constant 3 from the equation before integration.

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

substitute in:

= 3/2 integral ( du/ u)

and now integrate it using the natural logs rule.

= 3/2 ln |x^2 +1| + C

hope that helps

Answer 4

Let u = x^2+1
Then du = 2x dx --> xdx = du/2
So integral becomes 3du/2u =1.5 du/u
You should be able to take it from here.