Question

\[Evaluate \int\limits \frac{x ^{3}}{\sqrt{x ^{2}+1}}\]

Answer 1

trig sub i think x=tan t

Answer 2

Put \(u=x^2+1 \implies x^2=u-1 \implies 2xdx=du.\) The integral becomes
\[\frac{1}{2} \int \frac{u-1}{\sqrt{u}} du= \cdots\]

Answer 3

\[u=\sqrt{x ^{2}+1}, u ^{2}=x ^{2}+1, udu=xdx,x ^{2}=u ^{2}-1\]

Answer 4

\[\int\limits_{}^{}(u ^{2}-1)udu/u=\int\limits_{}^{}u ^{2}-1=1/3u ^{3}-u\]