Question #3691

What is the integral of tan^-1(4x) dx ?

Expert's answer

<img src="/cgi-bin/mimetex.cgi?%5Cint%7B%5Ctan%5E%7B-1%7D%284x%29dx%7D%20=%20[%5Carctan%284x%29%20=%20u,%20du%20=%20%5Cfrac%7B4%7D%7B16x%5E2%20+%201%7D,%20x=%20dv,%20v%20=%20dx]%20=%20%5C%5C%20=%20x%20%5Carctan%284x%29%20-%20%5Cint%7B%5Cfrac%7B4xdx%7D%7B16x%5E2%20+%201%7D%7D%20=%20x%20%5Carctan%284x%29%20-%20%5Cint%7B%5Cfrac%7B2dx%5E2%7D%7B16x%5E2%20+%201%7D%7D%20=%20%5C%5C%20=%20x%20%5Carctan%284x%29%20-%20%5Cfrac%7B1%7D%7B8%7D%5Cln%2816x%5E2%20+%201%29" title="\int{\tan^{-1}(4x)dx} = [\arctan(4x) = u, du = \frac{4}{16x^2 + 1}, x= dv, v = dx] = \\ = x \arctan(4x) - \int{\frac{4xdx}{16x^2 + 1}} = x \arctan(4x) - \int{\frac{2dx^2}{16x^2 + 1}} = \\ = x \arctan(4x) - \frac{1}{8}\ln(16x^2 + 1)">

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