calc
posted by marsha .
also:
integral of tan^(1)y dy
how is integration of parts used in that?
You write:
arctan(y)dy = d[y arctan(y)] 
y d[arctan(y)]
Here we again have used the product rule:
d(fg) = f dg + g df
You then use that:
d[arctan(y)] = 1/(1+y^2) dy
So, the integral becomes:
y arctan(y)  Integral of y/(1+y^2) dy
And:
Integral of y/(1+y^2) dy =
1/2 Log[1+y^2] + const.
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