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# p5 yeap integration! watch

1. can anyone help me? im useless

integrate x+1 / (1-x^2) ^1/2 dx

in simple english
integrate x+1 divided by 1-xsquared to the half with respect to x!

anyone?
2. (Original post by kriztinae)
can anyone help me? im useless

integrate x+1 / (1-x^2) ^1/2 dx

in simple english
integrate x+1 divided by 1-xsquared to the half with respect to x!

anyone?
INT (x+1) / (1-x^2) ^1/2 dx

= INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
3. (Original post by It'sPhil...)
INT (x+1) / (1-x^2) ^1/2 dx

= INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
ok thanx
4. (Original post by It'sPhil...)
INT (x+1) / (1-x^2) ^1/2 dx

= INT x/(1-x^2)^1/2 + 1/(1-x^2)^1/2 dx

For the first one sub u = 1 - x^2 if you want or recognise derivative second one is just arcsinx
hey can anyone help me on this same question? im stuck integrating the first part
ok to put it simpler!
how do you integrate:
x/(1-x^2)1/2 dx

anyone please? any help much apreciated thanx
5. >how do you integrate:
>x/(1-x^2)1/2 dx

as mentioned b4 use u = 1-x^2

so du/dx = 2x

du/2 = x dx

now x dx is what u have for your numerator.....so

Int 1/(u^1/2) du/2

1/2 * Int u^(-1/2) du

now its easy

mrm.
6. (Original post by Mrm.)
>how do you integrate:
>x/(1-x^2)1/2 dx

as mentioned b4 use u = 1-x^2

so du/dx = 2x

du/2 = x dx

now x dx is what u have for your numerator.....so

Int 1/(u^1/2) du/2

1/2 * Int u^(-1/2) du

now its easy

mrm.
i dont see it
dont worry about it ill find another way! thanx anyway

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