Integration?

Question

Hi,

How do you integrate (2x^2)/(1+x^2)?

I appreciate any help. Thank you!

Answer 1

Hi dear Canty
{ as you didnt want to explain step by step , so i don't waste your time darling , on the other hand three of your answers are correct as well }
Here is your answer;
∫(2x^2)/(1+x^2) dx = 2( x- Arc tan(x)) + c
Or
2 ( x - tan^-1( x) + c { they are same}

Good Luck Darling

Answer 2

As you see the degree of the denominator is that of the numerator(i.e. both have x^2 as the highest power of the variable)
Now a little manipulation will do the trick. For integration with respect to x, let "S'denote the integration(sum) symbol.
Then,
S ax dx = a S x dx
so,
S [2x^2 / (1+ x^2)] dx = 2 S[x^2 / (1+x^2)] dx
the numerator can now be written as (x^2 +1) - 1
so the expression becomes, [(1+x^2) / (1+x^2)] - 1/(1+x^2)
So the integral becomes,
2.S [ 1 - 1/(1+x^2)] dx = 2{ x - S [1/(1+x^2)] dx} -----------(1)
We're about done,
Consider the arctangent function (i.e tan^-1)
if y = arctan x
x = tan y
then dx/dy = sec^2 y = 1 + tan^2y = 1 + x^2
then by the chain rule of differentiation, dx/dy = 1/(dy/dx), so;
dy/dx = 1/(1+x^2)

So working backwards the integral of 1/(1+x^2) is arctan x
Note that this is the problem we now face in (1), so simply,
2{ x - S [1/(1+x^2)] dx} = 2{ x - arctan x} + C
where C is an arbitrary constant of integration.

Hope this helps!

Answer 3

2(x^2)/(1+x^2)
= 2(X^2+1-1)/(1+x^2)
= 2 - 2/(1+x^2)
= 2x-2arctan(x)
integral of 2 is 2x and 2/(1+x^2) = 2 arctan(x)

for a method above divide and remove the x^2 part then there may be x part and constant part

x part shall be derivative of denominator here it is not there and constant part arctan
if you want me to explain the detailed steps please e-mail me
through contact

Answer 4

let x^2 =t
2xdx=dt

t/(t+1)=1- [1/(t+1)]

integral=2 [t-ln(t+1)]
=2x^2-2ln (x^2+1) +c

Answer 5

there are formulas for every form of integration in calculus

ans: 2(x - Arctan(x))