Asked by Anonymous

integral(8&4) y dy/(y^2-2y-3)

Answers

Answered by Count Iblis
int y dy/(y^2-2y-3) =

int 1/2 2y dy/(y^2-2y-3) =

int 1/2 (2y - 2 + 2) dy/(y^2-2y-3) =

1/2 Log(y^2 - 2 y - 3) +

int dy/(y^2-2y-3)

int dy/(y^2-2y-3) =

int dy/[(y-3)(y+1)]

1/[(y-3)(y+1)] =

1/4 [1/(y-3) - 1/(y+1)] ----->


int dy/(y^2-2y-3) =

1/4 Log[(y-3)/(y+1)]
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