Asked by Anonymous
integral(8&4) y dy/(y^2-2y-3)
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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)]
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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