Question

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

Answer 1

What have you tried so far?

Answer 2

whenever you see a radical, the first thing to try is to substitute it :
\[\large u^2 = x^2+1\]

Answer 3

i bet you have tried that already :)
it doesn't seem to work out smoothly hmm

Answer 4

yes

Answer 5

what about trig substitution ?

Answer 6

not tried

Answer 7

try it, looks it simplifies nicely u = tan

Answer 8

not getting

Answer 9

are you on words diet today hmm ;p

Answer 10

may be

Answer 11

cn u give me the exact solution

Answer 12

well i am on diet too >.<

Answer 13

oh so u r not going to help me

Answer 14

i can try to help if you atleast attempt something

Answer 15

even i don't know how to work it yet, so we need to work it together

Answer 16

i dont know about trig. substitution

Answer 17

how about hyperbolics ?

Answer 18

or something like this :

multiply numerator and denominator by x^2, pull out a x from the botom radical and substitute
\[\large u = 1+\frac{1}{x^2}\]

Answer 19

they are all the things i would try if that was my homework problem

Answer 20

notice that derivative of arcsinhx is 1/(x^2+1)^1/2

that might help in getting ideas on how to attempt too

Answer 21

the answer given is in log

Answer 22

Integration by parts perhaps may help

Answer 23

shinal what should i take as first function then

Answer 24

okay it took me a while
\[\int\limits \frac{ x ^{2} }{ (x + 1)^{\frac{ 1 }{ 2 }} }dx\]
can be expressed as
\[\int\limits (x)(\frac{ x }{ (x ^{2}+1)^{\frac{ 1 }{ 2 }} })\]
have you remembered
\[\int\limits udv = uv - \int\limits vdu\]
let's do that for the given expression
we let
\[u = x \]
\[du = dx\]
\[v = \frac{ x }{ (x ^{2}+1)^{\frac{ 1 }{ 2 }} }dx\]
or
\[dv = x(x ^{2}+1)^{-\frac{ 1 }{ 2 }}dx\]
integrate to have the value for v
\[\int\limits \frac{ dv }{ dx } = \frac{ 1 }{ 2 } \int\limits 2x(x ^{2}+1)^{-\frac{ 1 }{ 2 }}dx\]
\[v = \frac{ 1 }{ 2 }(\frac{ (x^2+1)^{\frac{ 1 }{ 2 }} }{ \frac{ 1 }{ 2 } })\]
\[v = (x^2 + 1)^{\frac{ 1 }{ 2 }}\]
let's substitute it to \[\int\limits udv = uv - \int\limits vdu\]
\[ = x(x^2+1)^{\frac{ 1 }{ 2 }} - \int\limits (x^2+1)^{\frac{ 1 }{ 2 }}dx\]
since \[\int\limits \sqrt{u^2+a^2} du = \ln \left[ \sqrt{u^2+a^2}+u \right] +c\]
then we'll apply that and we'll have
\[= x(x^2+1)^\frac{ 1 }{ 2 } - \ln \left| \sqrt{x^2+1}+x \right| + c\]

Answer 25

u still need help :O

Answer 26

no ,thanks