Simple Integral

  • Thread starter rwinston
  • Start date
  • #1
36
0
Hi

I am trying to integrate

[tex]x \sqrt{1+x^2}dx[/tex]

by parts...but it seems to involve trigonometric functions - is it possible to solve this integral without using trig functions?


Thx
 

Answers and Replies

  • #2
1,074
1
Why are you trying to use integration by parts? This is a very simple integral if you use an appropriate substitution.
 
  • #3
mathman
Science Advisor
8,006
518
The simple substitution is u2=1+x2.
 
  • #4
1,341
3
Mathman's solution is the easiest, but you probably could do it by parts.
 
  • #5
1,753
1
The simple substitution is u2=1+x2.
Might confuse him/her.

Just let u = radican.

u=1+x^2
 
  • #6
36
0
Hmm. Ok if u = 1 + x^2

[tex] du = 2x dx [/tex]

[tex]\int x \sqrt{1+x^2}dx = \frac{1}{2}\int \sqrt{1+x^2}2xdx[/tex]

[tex]=\frac{1}{2} \int\sqrt{u}du[/tex]

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

But if I take u^2=1+x^2

[tex]u = \sqrt{1+x^2}[/tex]

[tex]du = \frac{x}{\sqrt{1+x^2}}dx[/tex]

[tex]\int{x\sqrt{1+x^2}dx} = \sqrt{1+x^2}\int{\sqrt{1+x^2}\frac{x}{\sqrt{1+x^2}}dx}[/tex]

[tex]=\sqrt{1+x^2}\int{u du}[/tex]

[tex]=\sqrt{1+x^2} \frac{{\sqrt{1+x^2}}^2}{2}[/tex]

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

I must be making a mistake somwhere...
 
  • #7
1,753
1
Hmm. Ok if u = 1 + x^2

[tex] du = 2x dx [/tex]

[tex]\int x \sqrt{1+x^2}dx = \frac{1}{2}\int \sqrt{1+x^2}2xdx[/tex]

[tex]=\frac{1}{2} \int\sqrt{u}du[/tex]

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

But if I take u^2=1+x^2

[tex]u = \sqrt{1+x^2}[/tex]

[tex]du = \frac{x}{\sqrt{1+x^2}}dx[/tex]

[tex]\int{x\sqrt{1+x^2}dx} = \sqrt{1+x^2}\int{\sqrt{1+x^2}\frac{x}{\sqrt{1+x^2}}dx}[/tex]

[tex]=\sqrt{1+x^2}\int{u du}[/tex]

[tex]=\sqrt{1+x^2} \frac{{\sqrt{1+x^2}}^2}{2}[/tex]

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

I must be making a mistake somwhere...
No, he meant to use u^2

So ...

[tex]u^2=1+x^2 \rightarrow udu=xdx[/tex]

And remember that you can only pull out constants:

[tex]c\int f(x)dx[/tex]
 
  • #8
412
4
wow, substitutions are so much mess. Try this way:

x.sqrt(1+x^2)

think about sqrt(1+x^2)
it has integral
(1+x^2)^3/2 --- [ref 1]

it has derivative 3x (1+x^2)^0.5

But, in original thing we only have x.sqrt(1+x^2)

so, you need 1/3 with [ref 1]
 

Related Threads on Simple Integral

  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
4
Views
628
  • Last Post
Replies
7
Views
941
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
853
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
841
  • Last Post
Replies
3
Views
823
Top