`intsqrt((a+x)/(a-x)) dx`

Answer

N/a

Explanation

Let `I = intsqrt((a+x)/(a-x))dx`
Put `x = acos2 theta`
`rArr dx = -a.sin2theta.2.d theta`
`:. I = -2intsqrt((a+acos2theta)/(a-acos2 theta)).a sin2 theta d theta`
`[:' cos2 theta = x/a rArr 2 theta = cos^(-1)'(x)/(a) rArr theta = 1/2 cos^(-1)'x/a]`
` =-2aintsqrt((1+cos2theta)/(1-cos2theta))sin2theta d theta = -2a intsqrt((2cos^(2)theta)/(2sin^(2)theta))sin2theta d theta`
`=-2aintcottheta.2sin2theta d theta= -2aint(costheta)/(sin theta).2sin theta.cos theta d theta`
` = -4aintcos^(2) theta d theta = -2a int(1+cos2theta)d theta`
`= -2a[theta+(sin2theta)/(2)]+C`
` =-2a[1/2 cos^(-1)'(x)/(a)+1/2sqrt(1-(x^(2))/(a^(2)))]+C`
` = -a[cos^(-1)(x/a)+sqrt(1-(x^(2))/(a^(2)))]+C`
Alternate Method
Let `I = intsqrt((a+x)/(a-x))dx = intsqrt(((a+x)(a+x))/((a-x)(a+x)))dx`
` = int((a+x))/(sqrt(a^(2)-x^(2)))dx`
`I = int(a)/(sqrt(a^(2)-x^(2)))+int(x)/(sqrt(a^(2)-x^(2)))dx`
`:. I = I_(1)+I_(2)`
Now, `I_(1) = int(a)/(sqrt(a^(2)- x^(2)))= asin^(-1)(x/a)+C_(1)`
and `I_(2)= int(x)/(sqrt(a^(2)-x^(2)))dx`
Put `a^(2) - x^(2) =t^(2) rArr -2x dx = 2t dt`
`:. I_(2) - int'(t)/(t)dt e = -int1dt`
`= -t + C_(2) = -sqrt(a^(2)-x^(2)) + C_(2)`
`:. I = asin^(-1)(x/a)+C_(1)-sqrt(a_(2)-x^(2)+C_(2)), [:' t^(2) = a^(2)-x^(2)]`
`I = asin^(-1)(x/a)-sqrt(a^(2)-x^(2))+C ,[:' C = C_(1)+ C_(2)]`