Question

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

Answer 1

Have you tried trig sub?

Answer 2

first you have to divide the numerator by denominator and then integrate.

Answer 3

i tried but it didn't solve it

Answer 4

show your work, please, follow surjithayer's instruction.

Answer 5

adjusting the numerator,
x^2 = (1/3)(3x^2) = (1/3) [ (3x^2 +1) -1 ]
then you can separate out the numerator and denominator...

Answer 6

\[\frac{ 1 }{ 3 }\int\limits\limits_{}^{} \frac{ {[3x^{2}+1}]-1 }{ 3x ^{2}+1 }dx\]

then
\[\frac{ 1 }{ 3 } \int\limits_{}^{}1dx -\frac{ 1 }{ 3 }\int\limits_{}^{}\frac{ 1 }{ 3x ^{2}+1 }\]
if it right then what i should do in the second part
\[\frac{ 1 }{ 3 }\int\limits_{}^{}\frac{ 1 }{ 3x ^{2}+1 }dx\]
because i stop here

Answer 7

correct till there,
now bring the denominator in the form x^2+a^2
like 3x^2+1 = (1/3) [x^2 + 1/3] = (1/3) [x^2 + (1/sqrt 3)^2]

since we have the form, x^2+a^2, we can put x= a tan t
here, x = tan t/ sqrt 3

Answer 8

look what if we from the step let u=3x^2+1 then x^2=(u-1)/3
ok i think it easier

Answer 9

letting u=3x^2+1 won't help.
as du will be 6x dx
and you don't have 'x' in numerator....

Answer 10

when we let u the euqation will be \[\frac{ \frac{ u-1 }{ 3 } }{ u }\] then \[\frac{ { u-1 } }{ 3u }\] then integrate

Answer 11

you still have dx part left to calculate...

Answer 12

u-1/3u dx <<<<<this dx needs to be calculated in terms of du

Answer 13

u=3x^2+1
du = 6x dx
dx = du/6x = du/ (6\sqrt (u-1)/3)
which makes it much more complicated...

Answer 14

splitting and making the trigonometric substitution is the easiest way for this...

Answer 15

thats right my mistake
thank you

Answer 16

no problem, ask me if you get stuck while doing trig. substitution.....or need any kind of help with this :)

Answer 17

oh and since you're new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 18

thank you