# Simple Integration by Parts

1. Jul 23, 2008

### irok

1. The problem statement, all variables and given/known data
#1.
Use Integration by parts to evaluate the integral
$$\int 2x \ln(2x) dx$$

#2.
Use Integration by parts to evaluate the integral
$$\int (\ln(3x))^{2} dx$$

#3.
Use Integration by parts to evaluate the integral
$$\int x e^{4x} dx$$

#4.
Evaluate the indefinite integral.
$$\int \sin(3x) \sin(11x) dx$$

3. The attempt at a solution
#1.
$$\int fg'\ = fg - \int f'g\$$
f=2x f'=2
g'=ln(2x) g=1/2x
$$\int fg'\ = \frac {2x}{2x}\ - \int \frac{2}{2x}\\$$
= 1 - ln|x| + C
I think I went wrong finding the anti-derivative of ln(2x). Would g(x) be 1/2 * 1/2x instead?

#2.
$$\int (\ln(3x))^{2} dx$$
Make both f(x) and g'(x) = ln(3x)?
I think I'm getting this question wrong because I don't know the anti-derivative of ln(3x).

#3.
$$\int x e^{4x} dx$$
f=x f'=1
g'=e^4x g = 1/4 e^4x
= [1/4 * x * e^4x - 1/4 * e^4x] + C

#4.
$$\int \sin(3x) \sin(11x) dx$$
$$1/2 \int \cos(3x-11x) - \cos(3x+11x) dx$$
$$1/2 \int \cos(-8x) - \cos(14x) dx$$
$$1/2 [\sin(-8x) - \sin(14x)] dx$$

I'm not sure what I'm doing wrong here.
Does sin(-8x) - sin(14x) = -sin(6x)?

2. Jul 23, 2008

### rootX

#1

d/dx (x^2.ln(x)) = 2x.lnx + x

integrate both things, and I find this way the easiest

#2

you would find anti-derivative for ln(3x) from the integration table. You probably will have that during your exam too

#3

I forgot the technique but I don't think your one is right, differentiate it
the answer should be '1/16*(-1+4*x)*exp(4*x)' figure out how to reach there ...

#4

$$1/2 \int \cos(-8x) - \cos(14x) dx [tex] is not equal to your final answer. You forgot to take care of constants. and cos(-a) = cos(a) "Does sin(-8x) - sin(14x) = -sin(6x)? " x = 1 does Does sin(-8*1) - sin(14*1) = -sin(6*1)?? Last edited: Jul 23, 2008 3. Jul 23, 2008 ### irok #1. Not sure how that way works. But can you confirm if the following are correct: f=2x f'=2 g'=ln(2x) g=1/(2x) #2. The anti-derivative for ln(3x) = 1/3x? f=ln(3x) f'=(1/x) g'=ln(3x) g=? #3. SOLVED I did something wrong when first trying this question. I didn't integrate e^4x properly so i missed the 1/4. so [tex]\frac {(4x-1)*e^{4x}}{16}$$ was my final answer after simplifying from $$\frac {x}{4} e^{4x} - \frac {1}{16} e^{4x}$$.
Thanks rootX.

#4. SOLVED
I totally forgot about taking care of the constants. Thank you rootX. Final answer is 1/2 * 1/8 [sin(8x)] - 1/2 * 1/14 [sin(14x)].

Last edited: Jul 23, 2008
4. Jul 23, 2008

### rootX

no.. d/dx(ln(x)) = 1/x not the integral check your integration table

Editing over.

5. Jul 23, 2008

### irok

#5.
Use integration by parts to evaluate the definite integral.
$$\int t e^{-t} dt$$

Is the following correct?
f = t f' = 1
g' = $$e^{-t}$$ g = $$-e^{-t}$$

$$-t e^{-t} ]^{1}_{0} - [e^{-t}]^{1}_{0}$$

Solved Thanks Rocomath.

Last edited: Jul 23, 2008
6. Jul 23, 2008

### rocomath

Yes, keep going!

Also, stick to "t", you used x for f.

7. Jul 23, 2008

### rootX

http://math2.org/math/integrals/more/ln.htm

I don't think you need to know that.

8. Jul 23, 2008

### irok

Cool, at first I didn't understand why it didn't matter if ln(2x) or ln(kx). Now I'm all cleared up! Thank you again rootX

9. Jul 23, 2008

### rootX

I thought you are messing derivatives with integration

as int(ln(x)) = 1/x ...

10. Jul 23, 2008

### irok

#2.
$$\int (ln(3x))^{2} dx$$

I'm still stuck on this one.

Do I use $$\int ln(3x) ln(3x)$$ and f=ln(3x) g'=ln(3x)

or

$$\int (ln(3x))^{2}$$ and f=ln(3x)^{2} g'=1

11. Jul 23, 2008

### rootX

try

f = [ln(3x)]^2 and g'=x

d/dx (x.[ln(3x)]^2 )= [ln(3x)]^2 + ..

I am not familar with that your notations, I never tried to learn them

12. Jul 23, 2008

### irok

I ended up with $$x(ln3x)^{2} - 2\int ln(3x)$$

u = $$(ln(3x))^{2}$$ du = $$2 ln(3x) \frac {1}{x}$$

Not sure how to integrate ln(3x). I know integral of ln(x) is xln(x)-x+C. Still not sure what to do when it's ln(3x). I am now assuming any integration of ln(kx) is 1/x

Last edited: Jul 23, 2008
13. Jul 23, 2008

### Defennder

$$\ln A + \ln B = \ln (AB)$$