Integration problems

  • Thread starter mad
  • Start date
  • #1
mad
65
0
Hello everyone.
I have been trying to do those 2 exercices for a while now and I can't get it.. We just started doing Integration by parts (is that how you call it in english?)
here are the problems

1) [tex] \int{ (x^3)(e^{x^2})} [/tex]
and
2) [tex] \int{ \frac{{(x)(e^x)}} { (x+1)^2}} [/tex]


for the 2nd one I tried
u=e^(x^2) dv= x^3
du= 2xe^(x^2) v= (x^4)/4

From there I just get stuck..
For #2, I dont even know how to start it because there are 3 terms.. Can you do that by parts?
I know these are rookies question, but please explain me
thanks :)
 
Last edited:

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,778
11
1) Try your luck with u = x² and dv = xe^x² (you can find v by substitution with x² = w)
 
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
4,778
11
2) Set

[tex]u = (x+1)^2 \ \ \ dv = xe^x dx \Rightarrow du = (2x^2 + 4x +2)dx \ \ \ v = \int(xe^x)dx[/tex]

Let us find v by integrating by parts, using

[tex]w = x \ \ \ dz = e^x dx \Rightarrow dw = dx \ \ \ z = e^x \Rightarrow v = xe^x - \int e^x dx = xe^x - e^x [/tex]

[tex]v = xe^x - e^x[/tex]

Continue. You will have to do many other integrations by parts to get to the final answer.
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
12,997
548
The last one is tricky.

[tex] \int\frac{x}{(x+1)^{2}} \ dx=\int\frac{(x+1)-1}{(x+1)^{2}} \ dx=\int \frac{dx}{x+1} -\int \frac{d(x+1)}{(x+1)^{2}} =\ln (x+1)+\frac{1}{x+1} [/tex]

Now

[tex] I:=\int e^{x}\frac{x}{(x+1)^{2}} \ dx [/tex]

can be part integrated to get

[tex] I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\int e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right) \ dx=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)[/tex]

[tex] -\int e^{x} \ln (x+1) \ dx-\int \frac{e^{x}}{x+1} \ dx [/tex]

Use part integration for the first of the last 2 integrals

[tex] I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\left[e^{x}\ln(x+1)\right]+\int \frac{e^{x}}{x+1} \ dx-\int \frac{e^{x}}{x+1} \ dx =\frac{e^{x}}{x+1} + C [/tex]

Daniel.
 
Last edited:
  • #5
quasar987
Science Advisor
Homework Helper
Gold Member
4,778
11
Dexter: What does "Docendo discitur" mean? Who's Seneca?
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
12,997
548
I'll let u discover who Seneca was.U can learn more from the internet.

"You learn by teaching others".It should be a motto for these forums...

Daniel.
 

Related Threads on Integration problems

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
898
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
474
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
886
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
984
  • Last Post
Replies
7
Views
1K
Top