Question

Integral. algebraic substitution

Answer 1

@zepdrix u israelii

Answer 2

Trig sub would work ok here.
What is algebraic sub?
They want you to do u=x^2+4 maybe?

Answer 3

\[\Large\rm \int\limits x^3 \sqrt{x^2+4}dx=\int\limits x^2\sqrt{x^2+4}(x~dx)\]

Answer 4

Looks like you've got almost everything you need.
Just need to solve for x^2 in that first substitution you established.

Answer 5

How about \(x=\sqrt u\), so that \(x^2=u\) and \(dx=\dfrac{du}{2\sqrt u}\), then
\[\begin{align*}\int x^3\sqrt{x^2+4}~dx&=\int u^{3/2}\sqrt{u+4}\frac{du}{2\sqrt u}\\
&=\frac{1}{2}\int u\sqrt{u+4}~du\\
&=\frac{1}{2}\int(t-4)\sqrt t~dt&\text{subbing }t=u+4,\text{ so }dt=du
\end{align*}\]

Answer 6

try what zepdrix # has written above i should be solved