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Solve and Explain Three Integrals

∫ (x^5)/[(1+x^3)^(3/2)] dx
∫ {square root of [(x+3)/(x+1)]}dx
∫ (cot^3 v)[(csc v)^(3/2)] dx

(I will use the $ sign for the integral sign)

Problem #1:
$ (x^5)/[(1+x^3)^(3/2)] dx

the power 3/2 in the denominator is throwing me off greatly, as is
the greater power (x^5) in the numerator.

attempt 1: let u=x^2 (so u^(1/2)=x) and du=2xdx the integral becomes
1/2$ [(u^2*u^(1/2))/(1+u*u^(1/2))^(3/2)]du
= 1/2$ [(u^3/2)/(1+u^(3/2))^(3/2)]du

at this point I wasn't sure of where to go so I restarted the
problem using intergration by parts

attempt 2: let u=1/[(1+x^3)^3] and du= {-9x^2/[(1+x^3)^4]dx let dv=
(x^5)dx and v= 1/6(x^6)
the integral was then
=1/6(x^6)*[1/(x+1)^3] - $1/6(x^6)*(-9x^2/[(1+x^3)^4])dx
=(x^6)/[6(x+3)^3]) - (9/6)$ x^2/[(x+3)^4]dx
but I couldn't figure what to do with the x^2


$ {square root of [(x+3)/(x+1)]}dx

I can't get around the square root for the entire problem. I can't
figure out the correct substitution method.

At first I tried to ignore the squareroot and integrate by part like
let u=x+3 and du=dx so dv=[1/(x+1)]dx and v=ln(x+1) but it didn't
seem to work correctly.
I thought maybe let u=(x+3)/(x+1) so the $u^(1/2) but the du
wouldn't work out correctly. I thought of partial fraction
integration but didn't know how to do that with the squareroot.

$ (cot^3 v)[(csc v)^(3/2)] dx

Once again, the fraction exponent is throwing me off. I know there
are substitutions and spliting of integral, but I can't get rid of
the (csc v)^(3/2)

${(cot^3 v)[(csc v)^(3/2)]}dv
${(cot^2 v)(cot v)[(csc v)^(3/2)]} dv cot^2 v= csc^2 v -1
${(csc^2 v -1)(cot v)[(csc v)^(3/2)]} dv
${(csc^2 v)(cot v)[(csc v)^(3/2)]} dv - ${(cot v)[(csc v)^(3/2)]}dv
I know that I can use trig sub. for the csc^2 v and cot, but I don't
know how to get rid of the csc v^(3/2) or even how to just get it to
be a csc v.

Any information that could help me with any of these three problems
would be extremely appreciated.

Solution Preview


Solution of three calculus problems is attached.

I have give file names in this way ...

Solution Summary

Three integrals are solved and explained. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.