How do I start this integration?

Question

⌠ln [(e^3x) – 1] dx
⌡ e^2x

* That's the natural log of the quantity ("e" to the 3x) minus 1 all over "e" to the 2x.

The only hint given was that it's POSSIBLE we will have to use ALL the Integration Techniques we've learned thus far. They are:
1. Regular u-substitution
2. Integration by Parts
3. Trig integrals
4. Partial Fraction Decomposition
5. Using Pythagorean Theorem w/ Right Triangles
6. Completing the square

I started with a regular u-sub. Letting u = e^3x - 1 and du = 3e^3x, but ended up at a dead end. 0_o Any feedback is greatly appreciated. :)

Answer 1

The maths teacher must really hate this class!

1. Make a substitution. Put u = e^x
du = e^x dx ie dx = du/u

2. The integral becomes

integral {ln(u^3-1)/u^2 du/u } or integral ln(u^3-1)/u^3 du

3. Integrate by parts - integral (udv) = uv - integral (vdu)

take u = ln(u^3-1) and dv = u^-3

this gives

-1/2 u^-2 ln(u^3-1) + integral { 1/2 u^-2 1/(u^3-1) 3u^2 du or

-1/2u^-2(u^3-1) +3/2 integral {1/(u^3-1)} du

4. The first part is fine and you can back substitute. Consider the second integral.

5. 1/(u^3-1) is a rational function and is resolved by partial fractions.

1/(u^3-1) = a/(u+1) + (bu +c)/(u^2 +u + 1) Noting u^3-1 =(u-1)(u^2+u+1)

This leads to

1 =a(u^2+u+1)+(bu+c)(u-1)

Expanding, collecting like terms and equating co-efficients yields a=1/3 b=-1/3 and c=-2/3

Therefore

1/(u^3-1) = 1/3 1/(u-1) - 1/3 (u+2)/(u^2+ u +1)

The first term is now integratable

integral {1/ (u-1) du} = ln(u-1).

Second term (ignoring the -1/3 for the time being)

(u+2)/(u^2+u+1) = 1/2(2u+1)/(u^2+u+1)+3/2(u^2+u+1)

First above is integratable note of form dz/z

Second term is as follows

1/(u^2+u+1) = 1/((u+1/2)^2 +3/4)

This is now of the form 1/(z^2+a^2) and its integral is

1/a tan^-1(z/a)

You have enough to fully intrgrate all parts. Just need to back substitute to recover x. I'll leave that to you.

PS Check my algebra as I did this on the run.

Answer 2

I must be interpreting the problem differently from the previous answerer, because I'm not seeing that there's a rational expression to separate. The ln is applied to [e^(3x)-1], and the e^(2x) is not included in that.

I would suggest making the substitution u=e^x. Then try integration by parts, which in turn leads to partial fractions. That part is going to get ugly because there will be an irreducible quadratic factor, but at least it's doable.

Answer 3

I think u should start by parts, by integrating e^-2x to have this integral:-0.5 * ln[e^(3x)-1]de-(2x), and complete by parts

Answer 4

first separate the rational expression with common

denominator of e^2x.

Simplify: ln(e^3x) = 3x (inverse property).

Now integrate. It is much easier!

Answer 5

In antiderivative technique ,we divide the area of any curve into countless form of small areas and discover the area of one such small area.Integrating the area of the small area from one particular shrink to a distinctive,we get the area of the curve. Integration provides the area of any form of curve.

Answer 6

1: by part: let u=ln(exp(3x)-1) ..... dv=dx/exp(2x)
this leads to
du=[1/(exp(3x)-1)].exp(3x).3.dx
v= -(1/2).(1/exp(2x)
then vdu=(-3/2). exp(x)/[exp(3x)-1].dx
now subistitute y=exp(x) dy=exp(x)dx
reduce vdu=(-3/2).dy/(y^3-1)
which ( i suppose) can be integrated by partial fraction