Question

Integrate?

`(2x+1)/(sqrt(1-x^2))dx`

Answer 1

`int(2x+1)/sqrt(1-x^2)dx` = `int (2x)/sqrt(1-x^2)dx + int 1/sqrt(1-x^2)dx`

`u = 1-x^2`
`du = -2x dx`
`-int u^(-1/2)du + arcsinx + c= -u^(1/2)/(1/2) + arcsinx + c = -2sqrt(1-x^2) + arcsinx + c`

Explanation:

You need to separate the integral first, then do a u-substitution on the side that has a 2x in the numerator and you need to use the arcsin integral trig formula to get the side with a 1 in the numerator.

Answer 2

Please look below.

Explanation:

`I = int` `(2x+1)/(sqrt(1-x^2))` `dx`

`= int` `(2x)/(sqrt(1-x^2)) + 1/(sqrt(1-x^2))` `dx`

`= arcsin(x )+ int` `(2x)/(sqrt(1-x^2))` `dx`

`J = int` `(2x)/(sqrt(1-x^2))` `dx`

Let:

`sin(u)= x`
`cos(u)` `du` = `dx`

`J = int` `(2sin(u)cos(u))/(sqrt(1-sin^2(u)))` `du`

`= int` `(2sin(u)cos(u))/(sqrt(cos^2(u)))` `du`

`= int` `2sin(u)` `du`

`= -2cos(u) + C`

`= -2cos(arcsin(x)) + C`

Therefore:

`int` `(2x+1)/(sqrt(1-x^2))` `dx = arcsin(x) - 2cos(arcsin(x)) + C`

This can be simplified further:

Let `theta = arcsin(x)`:
`sin(theta) = x`
`cos(theta) = sqrt(1-x^2)`

`cos(arcsin(x)) = sqrt(1-x^2)`

Finally:

`int` `(2x+1)/(sqrt(1-x^2))` `dx = arcsin(x) - 2sqrt(1-x^2) + C`