`intlog(sqrt(1-x)+sqrt(1+x))dx` का मान ज्ञात कीजिए ।

Explanation

`intlog(sqrt(1-x)+sqrt(1+x))dx`
`=int1.log(sqrt(1-x)+sqrt(1+x))dx`
`=x log(sqrt(1-x)+sqrt(1+x))-int(x)/(sqrt(1-x)+sqrt(1+x)) (1/(-2sqrt(1-x))+1/(2sqrt(1-x)))dx`
`=x log(sqrt(1-x)+sqrt(1+x))-intx/(sqrt(1-x)+sqrt(1+x))1/2(sqrt(1-x)-sqrt(1+x))/sqrt(1-x^2)dx`
`=xlog(sqrt(1-x)+sqrt(1+x))-1/2int(x)/sqrt(1-x^2)((sqrt(1-x)-sqrt(1+x))^2)/((-2x))dx`
`=xlog(sqrt(1-x)+sqrt(1-x))=1/4int(2-2sqrt(1-x^2))/sqrt(1-x^2)dx`
`=x log (sqrt(1-x)+sqrt(1+x))=1/2int(dx)/(sqrt(1-x^2))dx`
`=x log(sqrt(1-x)+sqrt(1+x))+1/2sin^(-1)x-1/2x+c`