Question

May someone help me with this integral?

Answer 1

this is the answer

Answer 2

\[u=(2-x^2)\]\[du=-2xdx\]\[=-x^2x(2-x^2)dx\]\[{-\int\limits_{}^{}}^{}(2-u)u^{12}du\] \[{\int\limits_{}^{}}^{}(u-2)u^{12}du\] \[{\int\limits_{}^{}}^{}(u^{13}-2u^{12})du\]\[ \frac{ u^{14} }{ 14 } -\frac{ u^{15} }{ 15 } \]
resubstitute u with 2-x^2 and it is the result..