help integral calculus?

Question

ln(x^3-3x-1)^5
--------------------dx
(x+1)(x-1)

(x^2-3x)
------------dx
(2-4x)

5
-------ds
3-4^s

e^(3x)-2e^(2x)+4e^x-2
------------------------------...
1-e^x

e^[ln(5x-3)^2]
--------------------dx
5x-3

w(3w-2)(1+lnw)
-----------------------dw
w+lnw^w

Answer 1

These are tough! Let's try to do them one at a time
1). Recall that ln a^r = r log a.
So we can put the 5 in front
to get 5∫ ln(x³-3x-1) dx/(x²-1) .
Now let's use partial fractions on the denominator. We get
-5/2*[ ∫ ln(x³-3x-1) dx /(x+1) - ∫ln(x³-3x-1) dx /(x-1)]
Unfortunately, here we run out of luck:
∫ ln(P(x)) dx /(x+1) and ∫ ln(P(x)) dx/(x-1),
where P(x) is a polynomial, are both non-elementary.
In fact, both involve the dilogarithm function.

2. This is a rational function and the degree
of the numerator is larger than the degree
of the denominator. So do a long division to get
∫[( -1/4 x+ 5/8) + (5/4)/(2-4x)] dx
Now break the integral in 2 and carry on from here.
I'll let you finish it!

3. Let u = 4^s, ln u = s ln 4, s = ln u/ln 4, ds = du/ ( u ln 4)
We get
(5/ ln 4)*∫ du/u(3-u).
Now use partial fractions again, to get
5/(3 ln 4)*∫ [1/u - 1/(u-3) ] du
5/(3 ln 4)* [ ln u - ln(u-3)] + C =
5/ln 64*[ s ln 4 - ln(4^s-3)] + C

4. Let u = e^x, x = ln u, dx = du/u
So you get
∫ (u³-2u² +4u-2) du / u(1-u)
Again, the degree of the numerator exceeds
that of the denominator, so do a long division
and proceed as in no.2.

5. Recall that e^ln(u) = u, so your integral is just
∫ (5x-3)² dx/ (5x-3) = ∫ (5x-3) dx
= 5x²/2 - 3x + C.

6. I will assume that the last term of the
denominator is ln(w^w), else the problem is hopeless.
Now recall ln(w^w) = w ln w.
So your fraction is
w(1+ ln w)(3w-2)/w(1+ln w) = 3w-2.
and the integral is
3w²/2 -2w + C.

What a collection of problems!
Anyway, hope that helps a bit.