∫ sec^2 x tanx dx ------------------- (1+sec^2 x) ^2

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∫ sec^2 x tanx dx ------------------- (1+sec^2 x) ^2

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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are you sure it's over \[(1+\sec^2x)^2\]?
and not : \[(\sec^2x - 1)^2?\]
if not, then you can do it this way ^_^: 1) substitute sec^2x in the denominator with (1+tan^2x) and you'll get : \[\int\limits_{}^{}(\sec^2xtanx)/(1+1+\tan^2x)^2dx\] \[=\int\limits_{}^{}(\sec^2xtanx)/(2+\tan^2x)^2dx\] 2) let u = tanx and du = sex^2x dx: \[=\int\limits_{} (u)/(2+u^2)^2 du\] get (2+u^2)^2 up and you'll have : \[=\int\limits_{} u(2+u^2)^{-2}du\] now you can use integration by parts by letting the following as given : - f' = (2+u^2)^(-2) and g = u find f and g' then use the general equation which is: \[\int\limits_{}f'gdx = fg - \int\limits_{}fg'dx\] or partial fractions to solve this integration :) I hope I have given you hint, you can take it from here ^_^

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