a. Integral (x^2)/(sqrt(1+(x^2)))

Question

Would I separate these two into 2 separate integrals? Like: Integral of x^2 and the other integral of 1/sqrt(1+(x^2))

b. Integral (x^7)/(ln(x^4))dx
Do I use integration by parts for this?
I put u= lnx
du = 1/x
dv = x^7
v = (x^8)/8
It doesn't seem to work though.

Answer 1

Also, how would I do the first problem without sinhx?

Answer 2

(a) would you separate 6^2/√(6^2+1) like that? I think not.

use trig substitutions. Let
x = tanθ
1+x^2 = sec^2θ
dx = sec^2θ dθ
and you have
∫tan^2θ/secθ (sec^2θ dθ)
= ∫tan^2θ secθ dθ
= ∫secθ secθtanθ dθ
Now let u = secθ and you have
∫u du

(b) let
u = x^4
du = 4x^3 dx
and you have
∫x^4/ln(x^4) x^3 dx
= 1/4 ∫u/lnu du

That's not possible using elementary functions. I suspect a typo.

Answer 3

For part a, how do you know to convert that to tan and sec? Is there some formula?

Answer 4

Also, I got as my answer: 1/2sec^2(theta)
But the answer given is:
[1/2(x)(sqrt((x^2)+1))] - [1/2ln(x+(sqrt(1+(x^2))))] + C

Answer 5

you pick your substitutions because of identities

sin^2 + cos^2 = 1
cosh^2 = sinh^2 + 1

by dividing by sin^2 or cos^2 you can get the corresponding ones involving tan,sec,cot,csc, etc.

as for the answer, recall that the integral of sec is log|sec+tan|, and that the inverse hyperbolic functions also can be written as strange logs.

Answer 6

I'm glad you have such faith in my math, but I make mistakes like anyone else. Starting with

∫tan^2θ secθ dθ
I should have gone on to say
∫tan^2θ secθ dθ
∫sec^3θ - secθ dθ
Now, we know that secθ gives us log|secθ+tanθ| = log|x+√(1+x^2)|

For sec^3θ we use integration by parts
u = secθ
dv = sec^2θ dθ

du = secθtanθ dθ
v = tanθ, so we have

tanθsecθ - ∫secθtan^2θ dθ
now we're back to where we started, so we do it again and we wind up with your answer.