Question

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)

Answer 1

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

Answer 2

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 ><"

Answer 3

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.

Answer 4

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

Answer 5

ok

Answer 6

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
\[\int\limits_{}^{}u^{2} -1 du\]

Answer 7

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