Question

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

Answer 1

the top is almost the derivative of the bottom

Answer 2

since x^3 derives to 3x^2, id multiply by 3/3 and use what i need from that to ln this up

Answer 3

is it let u = 2+x^3
du= 3x^2
dx= 1/3 x^2 du ?

Answer 4

that would be fine as well

Answer 5

\[\frac{x^2}{3x^2~u}=\frac{1}{3}\frac1u\]

Answer 6

but then \[\int\limits \frac{ x^2 }{ u } * \frac{ du }{ 3x^2 } = \int\limits 1/3u * du\]

Answer 7

oh ok, i see what you're saying. so is the answer 1/3 LN (2+x^3) + c?

Answer 8

missing the absolute markers, but yes

Answer 9

or is it \[\frac{ 1 }{ 3}*\frac{ 1 }{ u } = \frac{ 1 }{ 3 } * \frac{ 1 }{ 2+x^3}= \frac{ 1 }{ 6+3x^3 }\]

Answer 10

\[\frac{x^2}{2+x^3}\]
\[\frac33\frac{x^2}{2+x^3}\]
\[\frac13\frac{3x^2}{2+x^3}\]
\[\frac13ln(2+x^3)\]

Answer 11

thank you so much! these integrals are not easy to me

Answer 12

practice helps :)