What are the steps for using Integration by Parts to evaluate an integral?

[irok]

Homework Statement

#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

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)?

 

[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

<br /> 1/2 \int \cos(-8x) - \cos(14x) dx<br /> &lt;br /&gt; &lt;br /&gt; is not equal to your final answer. You forgot to take care of constants. and cos(-a) = cos(a)&lt;br /&gt; &lt;br /&gt; &amp;quot;Does sin(-8x) - sin(14x) = -sin(6x)? &amp;quot;&lt;br /&gt; x = 1 &lt;br /&gt; does Does sin(-8*1) - sin(14*1) = -sin(6*1)??

 

[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 \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)].

 

[rootX]

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

Editing over.

 

[irok]

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

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.

 

[rocomath]

Yes, keep going!

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

 

[rootX]

#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=?

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

I don't think you need to know that.

 

[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

 

[rootX]

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

I thought you are messing derivatives with integration

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

 

[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

 

[rootX]

#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

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

 

[irok]

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

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

 

[Defennder]

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