Simple integral please help me

integral 3x^2 / x^2 + 1 dx

okay i think the answer to this one is obvious but i just cant put my finger on it. please help! if you can please break it down for me. thanks again

I assume your question is this:
integral {3(x^2)/[x^2 + 1] dx}

You can do this:
integral {(3(x^2)+3-3)/[x^2 + 1] dx}
= integral ({(3(x^2)+3)/[x^2 + 1]} dx - {3/[x^2 + 1]} dx})
= integral (3dx - {3/[x^2 + 1]}dx)

The first one can be done very easily. The integral value is just 3x
The second one is simply the integral of arctan(x). The integral value is just 3*arctan(x)

So the integral is: 3x + 3*arctan(x) + C, C = arbitrary constant

first you need to take the 3 out as it is a constant
This will be an arctan invoving other steps

3 integral x^2/x^2+1 dx


now you have to do long division because you can't split the numerator
1 - (1/x^2+1)
------------
x^2+0x+1 |x^2+0x+0
-( x^2+0x+1)
0 + 0 -1
-1

so 3 integral 1 -(1/x^2+1)

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

3x- 3(1/1 arctan x/1)

for the 2nd u=x a=1 a is the number

3x-3 arctan x

without simplifying that first it WOULD be a very tough integral. however, you forgot that you can divide x^2 by x^2! 3x^2/x^2=3 :). so the new problem is
integral [3+1] dx
integral [4] dx
4x

the answer is 4x
hope this helped

3x^2/(x^2+1)
=(3x^2+3)/(x^2+1) - 3/(x^2+1)
=3 - 3/(x^2+1)
Integrate: 3x - tan-1(x) + C

∫(3x²)/(x² + 1) dx

∫(3x² + 3 - 3)/(x² + 1) dx

∫3 - 3/(x² + 1) dx = 3x - 3*arctan(x) + C

I am not good with integral, but the answer is Ln|x^3+1|+C