integrate (x^2)/(1+x^2) dx?

Answer 1

because the power on the top equal the power on the bottom so
on the top i add 1 and minus 1( didnt matter right? because you add 1 and minus 1 so that is 0)
on the top: x^2 +1-1
(x^2+1)-1
ok now
[(x^2+1)-1]/(x^2+1)
(x^2+1)/(x^2+1) -1/(x^2+1)
(by separate the fraction rule)
so
1-1/(x^2+1)
take integral of that
int(1-1/(x^2+1)dx)
use the rule of antiderivative
so int(1dx)-int(1/(x^2+1))
x-arctanx+C
(arctanx = tan^-1x or inverse tanx)
solution is x-arctanx+C
good luck.

Answer 2

very easy question

just apply numerator/denominator=questient+remainder/denominator

answer will be x - tanx +c

there is one more method add subtract numeror 1 and solve

u will get

integration of 1 dx - integration of 1/x^2+1

which on integration give back

x- tanx + c

Answer 3

split into:
integral of{[(1+x^2)/(1+x^2)]dx} - integral of [dx/(1+x^2)]
=integral of dx- integral of[dx/1+x^2]
=x-tan inverse(x)

Answer 4

x^3/3+x^5/5+constant

Answer 5

And the answer is (envelope pls): x-arctanx

Have a nice weekend.....