An integral problem

  • #1
1,395
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i tried to solve these integral by substitution
in the first integral i substituted x^2 for t

in the second integral i substituted e^x

it didnt work

how to solve these integrals?

plz help
 

Answers and Replies

  • #2
1,395
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here is the file

here is the file
 

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  • #3
9
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mm i think you must use fraction partial ...and in denominator adding +1 and - 1 . to use fraction partial .

1/(x^4 +1 +1 -1 ) = 1/(x^4 - 1 )+2

1/[(x^2-1)(x^2+1)+2]

1/[(x-1)(x+1)((x^2 -1)+2) +2]

1/[(x-1)(x+1)((x-1)(x+1)+2)+2]

and you can use this web http://integrals.wolfram.com/index.jsp
 
  • #4
9
0
mm i think you must use fraction partial ...and in denominator adding +1 and - 1 . to use fraction partial .

1/(x^4 +1 +1 -1 ) = 1/(x^4 - 1 )+2

1/[(x^2-1)(x^2+1)+2]

1/[(x-1)(x+1)((x^2 -1)+2) +2]

1/[(x-1)(x+1)((x-1)(x+1)+2)+2]

and you can use this web http://integrals.wolfram.com/index.jsp

and

and respect to second probelm

int[1/(e^x - 1)]

put 1 = e^x/e^x

its become int[ e^x/(e^2x - e^x)]

put u = e^x subtitution

du = e^x dx

du = u dx

so dx = du/u

int [u du/(u^2 - u ) u]

and u =! 0 cuz u = e^x and exponantioal never will be zero

int[du/u(u-1)] its fraction partial
 
  • #5
Gib Z
Homework Helper
3,346
5
First one is not fun at all, try x.users way.

For your second one, its a simple substitution.

[tex]u=\sqrt{e^x-1}[/tex]

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

So [tex]\int \frac{1}{\sqrt{e^x-1}} dx[/tex] becomes

[tex]\int \frac{1}{u} \frac{2u}{u^2+1} du[/tex].

u's cancel out, take out the factor of 2.

[tex]2\int \frac{1}{1+u^2} du[/tex].

That integral is easy, arctan.

[tex]\int \frac{1}{\sqrt{e^x-1}} dx = 2\arctan (\sqrt{e^x-1})[/tex]
 
  • #6
VietDao29
Homework Helper
1,423
2
1/[(x-1)(x+1)((x-1)(x+1)+2)+2]

and you can use this web http://integrals.wolfram.com/index.jsp
How can you Partial Fraction this?
[tex]\frac{1}{(x - 1) (x + 1) ((x - 1)(x + 1) + 2) + 2}[/tex]? :confused:

For the first one, you can try the following:

[tex]\int \frac{dx}{x ^ 4 + 1} = \frac{1}{2} \int \frac{(x ^ 2 + 1) - (x ^ 2 - 1)}{x ^ 4 + 1} dx = \frac{1}{2} \left( \int \frac{x ^ 2 + 1}{x ^ 4 + 1} dx + \int \frac{x ^ 2 - 1}{x ^ 4 + 1} dx \right) = \frac{1}{2} (I_1 + I_2)[/tex]

[tex]I_1 = \int \frac{x ^ 2 + 1}{x ^ 4 + 1} dx = \int \frac{1 + \frac{1}{x ^ 2}}{x ^ 2 + \frac{1}{x ^ 2}} dx = \int \frac{1 + \frac{1}{x ^ 2}}{\left( x - \frac{1}{x} \right) ^ 2 + 2} dx[/tex]

Now, make the u-substitution: [tex]u = x - \frac{1}{x} \Rightarrow du = \left(1 + \frac{1}{x ^ 2} \right) dx[/tex], your integral will become:

[tex]I_1 = \int \frac{du}{u ^ 2 + (\sqrt{2}) ^ 2} = \frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) + C_1 = \frac{1}{\sqrt{2}} \arctan \left( \frac{x + \frac{1}{x}}{\sqrt{2}} \right) + C_1[/tex]

The second integral can be done by the u-substitution: [tex]u = x + \frac{1}{x}[/tex]

[tex]I_2 = \int \frac{x ^ 2 - 1}{x ^ 4 + 1} dx = \int \frac{1 - \frac{1}{x ^ 2}}{x ^ 2 + \frac{1}{x ^ 2}} dx = \int \frac{1 - \frac{1}{x ^ 2}}{\left( x + \frac{1}{x} \right) ^ 2 - 2} dx = ...[/tex]
Can you go from here? :)
 
Last edited:
  • #7
1,395
0
thank alot!!!
 

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