Question

can anyone can help me integrate this

Answer 1

\[e ^{\int\limits\limits_{}^{}\frac{ 2t^3 }{ t^2+1}dt}\]

Answer 2

are you trying to integrate from e ??
if yes then e to what ?

Answer 3

integrate the first then plug in the e \[\int\limits_{}^{}2t^3/t^2+1 (dt)\]

Answer 4

divide the numerator by denominator

Answer 5

then it is vey simple
check it

Answer 6

@Caerus

Answer 7

check the attachment /

Answer 8

@Caerus
when transcribing type-set expressions containing fractions to a single line, numerators and denominators must be enclosed in parentheses, as follows:
\(\int\limits_{}^{}2t^3/(t^2+1) (dt)\)
Otherwise a different expression will result.
The same is needed when using a calculator.
5/2+3 is not the same as 5/(2+3).

Answer 9

opps forgot about that
when the numerator exponents > denominator
use long division method

Answer 10

\(\color{black}{\displaystyle \int\limits\frac{ 2t^3 }{ t^2+1}{\rm d}t=\int\limits\limits_{}^{}2t\left(\frac{ t^2 }{ t^2+1}\right){\rm d}t}\)

\(\color{black}{\displaystyle {\rm u} =t^2+1}\)
\(\color{black}{\displaystyle {\rm u-1} =t^2}\)
\(\color{black}{\displaystyle {\rm d u} =2t{\tiny~}{\rm d}t}\)

\(\color{black}{\displaystyle =\int\limits\frac{ {\rm u}-1 }{\rm u}\rm du.}\)

Answer 11

Can you go from there? :)