t = y^2+1
2ydy = dt
∫ [( √(t-1)) / (t+1) dt =
∫y/(y^2+2) 2ydy =
2∫y^2/(y^2+2)dy = 2∫dy-4∫1/(y^2+2)dy =
2y - (4/√2)arctan(y/) +C
2√(t-1) -2√2arctan(√(t-1)/√2)+C
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Verified answer
t = y^2+1
2ydy = dt
∫ [( √(t-1)) / (t+1) dt =
∫y/(y^2+2) 2ydy =
2∫y^2/(y^2+2)dy = 2∫dy-4∫1/(y^2+2)dy =
2y - (4/√2)arctan(y/) +C
2√(t-1) -2√2arctan(√(t-1)/√2)+C