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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


Integral of y/(1+y^2) dy =

1/2 Log[1+y^2] + const.

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