help with calclulus

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help with calclulus

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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1 Attachment
can you explain to me what was done? I undertsnad what they did with the infinity but I dont get how they added a 2 to the integrand
1 Attachment
if you look at the intergrand they have put 1/2 out the front and the numerator is 2y...... remember 1/2 x 2y = y.....

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LOL but why didthey do that?
1 Attachment
sorry abt the attachment but there seems to be a bug
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its to make the numerator look like the derivative of the denominator.... which would suggest the solution would be a ln function
the solution might be... let u = y^2 du = 2y \[1/2\int\limits_{0}^{b} du/(u^2 + 1)\] which is tan^(-1)
oops should read 1/2 tan^(-1) (u) then re substitute 1/2 tan^(-1) (y^2)
Thanks that was clear campbell :D That is exactly what I wanted to know

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