Integration problem

  • Thread starter Zerkor
  • Start date
  • #1
18
0

Homework Statement


find the definite integral [itex]\int\frac{x^3}{\sqrt{x^2 + 1}}[/itex] dx from 0 to 1


Homework Equations





The Attempt at a Solution


This problem is in the integration by parts section .. I chose u = x^3 , and dv=[itex]\frac{1}{\sqrt{x^2 + 1}}[/itex] so v = [itex]\frac{x^4}{4}[/itex] and du = -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]) , so the integral is equal to [itex]\frac{x^4}{4}[/itex] . [itex]\frac{1}{√x^2 + 1}[/itex] - [itex]\int\frac{x^4}{4}[/itex] . -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]) .. and my problem is the integral on the right hand side of the equation; I don't know how to integrate it and I don't know whether if I've chosen the parts correctly or there is a better way of choosing the parts
 
Last edited by a moderator:

Answers and Replies

  • #2
hilbert2
Science Advisor
Insights Author
Gold Member
1,525
545
How about first doing a change of variables s=x2. Now ds=2xdx and the integrand simplifies considerably.
 
  • #3
STEMucator
Homework Helper
2,075
140
If the change of variable doesn't float your boat, you could use a trig substitution of ##x = atanθ##.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
It isn't the easiest way to work the problem, but if you really want to use integration by parts, try$$
u = x^2,\,dv=\frac x {\sqrt{x^2+1}}$$
 
  • #5
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
In the OP, if u = x^3, then du = 3x^2 dx, not that weird expression. This is basic differentiation.

And if dv = dx/SQRT(x^2+1), then v certainly is not equal to (1/4)*x^4.

You should review the formulas for derivatives and simple integrals.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
965

Homework Statement


find the definite integral [itex]\int\frac{x^3}{\sqrt{x^2 + 1}}[/itex] dx from 0 to 1


Homework Equations





The Attempt at a Solution


This problem is in the integration by parts section .. I chose u = x^3 , and dv=[itex]\frac{1}{\sqrt{x^2 + 1}}[/itex] so v = [itex]\frac{x^4}{4}[/itex] and du = -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]
You have "u" and "v" confused. With u= x^3, du= 3x^2dx- you differentiated v. With [itex]dv= \frac{1}{\sqrt{x^2+ 1}}[/itex], [itex]v= arctan(x)[/itex]- you integrated u.

, so the integral is equal to [itex]\frac{x^4}{4}[/itex] . [itex]\frac{1}{√x^2 + 1}[/itex] - [itex]\int\frac{x^4}{4}[/itex] . -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]) .. and my problem is the integral on the right hand side of the equation; I don't know how to integrate it and I don't know whether if I've chosen the parts correctly or there is a better way of choosing the parts

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #7
3
0
A one solve the definite integral

∫x2/√x2+1 dx={z=x2+1,dz=2*x*dx}=
∫(z-1)/(2√z) dz=1/3*z3/2-√z={x=0⇔z=1,x=1⇔z=2}=(2-√2)/3
 
  • #8
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
All these responses and suggestions, and Zerkor never returned to his thread. :frown:
 
  • #9
18
0
It isn't the easiest way to work the problem, but if you really want to use integration by parts, try$$
u = x^2,\,dv=\frac x {\sqrt{x^2+1}}$$

I tried this one and it didn't work .. In the sections I've studied in the previous two days I learned trigonometric substitution and miscellaneous substitution; I think I have to try working this problem with those techniques
 
  • #10
18
0
Thanks a lot guys for your help and I'm sorry for replying after a long period but I didn't access the forum through those two days :)
 
  • #11
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
Without showing your work, we can't say why integration by parts didn't work for you, except you must have made a mistake. IBP does work for this integral, but your earlier postings show that you do not have a grasp on how to apply it correctly.
 
  • #12
18
0
Without showing your work, we can't say why integration by parts didn't work for you, except you must have made a mistake. IBP does work for this integral, but your earlier postings show that you do not have a grasp on how to apply it correctly.

May be you are right. But I've just worked it using trigonometric substitution using x = tanθ and it worked correctly .. Maybe there exist a way to find it using IBP but it's a hard way and there is no use for it if there is an easier way :)
 
  • #13
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
May be you are right. But I've just worked it using trigonometric substitution using x = tanθ and it worked correctly .. Maybe there exist a way to find it using IBP but it's a hard way and there is no use for it if there is an easier way :)

But since you are learning calculus, don't you think it might be of some value to figure out why it's not working for you?
 

Related Threads on Integration problem

  • Last Post
Replies
2
Views
587
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
644
  • Last Post
Replies
5
Views
941
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
2
Views
855
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
5
Views
893
Top