Question #0c04e

1 Answer
Truong-Son N.
Oct 22, 2017

I got #arcsinx - sqrt(1-x^2) + C#, #" "" "x in [-1,1)#.

Well, you might as well multiply by the conjugate term to see what happens. I would choose this way, using #sqrt(1+x)#:

#int sqrt(1+x)/(sqrt(1-x)) cdot sqrt(1+x)/(sqrt(1+x))dx#

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

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

The first integral is known to be #arcsinx#, and the second is a u-substitution. Let:

#u = 1-x^2#
#du = -2xdx#

And so, we divide by #-2# to get:

#= arcsinx - 1/2 int (-2x)/sqrt(1-x^2)dx#

#= arcsinx - 1/2 int 1/sqrtudu#

#= arcsinx - 1/2 cdot 2sqrtu#

#= color(blue)(arcsinx - sqrt(1-x^2) + C)#, #" "" "x in [-1,1)#

To show that this works, the derivative should give the original integrand.

#d/(dx)[arcsinx - sqrt(1-x^2)]#

#= 1/sqrt(1-x^2) - 1/(2sqrt(1-x^2)) cdot -2x#

#= 1/sqrt(1-x^2) + x/(sqrt(1-x^2))#

#= (1+x)/sqrt(1-x^2)#

#= (sqrt(1+x)cancelsqrt(1+x))/(sqrt((1-x)cancel((1+x)))#

#= sqrt(1+x)/sqrt(1-x)# #color(blue)(sqrt"")#,

provided #x >= -1# and #x < 1#.

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