Question Number 102840 by Dwaipayan Shikari last updated on 11/Jul/20
Answered by OlafThorendsen last updated on 12/Jul/20
![y = Y^2 −x+1 2Y′Y−1+(√Y^2 ) = Y^2 +2 2Y′Y−1+Y = Y^2 +2 2Y′Y = Y^2 −Y+3 Y′((2Y)/(Y^2 −Y+3)) = 1 Y′[((2Y−1)/(Y^2 −Y+3))+(1/(Y^2 −Y+3))] = 1 Y′[((2Y−1)/(Y^2 −Y+3))+(1/((Y−(1/2))^2 +((11)/4)))] = 1 Y′[((2Y−1)/(Y^2 −Y+3))+((4/(11))/((4/(11))(Y−(1/2))^2 +1))] = 1 ln(Y^2 −Y+3)+(2/(√(11)))arctan(2/(√(11)))(Y−(1/2)) = x+C](Q102982.png)