The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

p5 yeap integration!

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?
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
Reply 2
Avatar for kriztinae
kriztinae
OP
14
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
Reply 3
Avatar for kriztinae
kriztinae
OP
14
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
>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.
Reply 5
Avatar for kriztinae
kriztinae
OP
14
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

Related discussions

Latest

Trending

Trending

Articles for you