anonymous 5 years ago Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways:? a) Integration by parts b) The substitution u = sqrt(x^2+1)

1. anonymous

b)$1/2\int\limits_{}^{}(u^{2}-1/u) du$

2. anonymous

i think i know part A ) $\int\limits x ^{3}\div \sqrt{x ^{2}+1}$ you can say that sqrt of x^2 + 1 is (x^2 + 1)^1/2 the answer will be > [x^4/4] / (x^2+1)^1/1 btw am not sure ><"

3. anonymous

a) $uv-\int\limits_{}^{}vdv$ it is a a very long problem, I wouldn't do it by parts and I could see your professor really want you to do it on HW, but it will unlikely on your test.

4. anonymous

hey dumbcow i just need to ask you one more question about the question i put up before. when your free lol:)

5. anonymous

ok

6. anonymous

i may be wrong, but i dont see how audia4 got the answer for part b) $u = \sqrt{x^{2}+1}$ $du =\frac{x}{\sqrt{x^{2}+1}} dx$ $\ dx=\frac{\sqrt{x^{2}+1}}{x} du$ replacing dx cancels out sqrt(x^2+1) and one x on top leaving x^2 du $x^{2} = u^{2} -1$ so integral becomes $\int\limits_{}^{}u^{2} -1 du$

7. anonymous

i was wrong on b), wasnt thinking, sorry guys! dumbcow got the right answer!!!!