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Evaluate the integral dy (tan-'[y/8)) (64+y?) ( 1 + dy (tan-'(4/8)) (64+y?) =

User Jtabuloc
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Answer: ln|y/8| + C
Explanation:

First, we need to recognize that the derivative of arctan(x) is 1/(1+x^2). Therefore, the derivative of arctan(y/8) is 8/(64+y^2).

Now, using the substitution u = y/8, we can rewrite the integral as:

∫(1/u)(64+64u^2)(8/(64+64u^2))du

Simplifying, we get:

∫(1/u)du = ln|u| = ln|y/8|

Therefore, the final answer is:

ln|y/8| + C

where C is the constant of integration.

User Ray Zhang
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