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\( \int \frac{\tan ^{-1} x}{1+\frac{1}{x^{2}}} d x \) |
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Answer» \(\int\frac{tan^{-1}x}{1+\frac1{x^2}}dx\) = \(\int\frac{x^2tan^{-1}x}{1+x^2}dx\) put tan-1 x = t ⇒ \(\frac1{1+x^2}dx=dt\) \(\therefore\) \(\int\)\(\cfrac{tan^{-1}x}{1+\frac1{x^2}}dx\) = \(\int t\,tan^2t dt\) = t\(\int tan^2 t dt\) - \(\int 1.(tan t - t)dt\) (\(\because\) \(\int\) tan2 t dt = \(\int(sec^2t-1)dt= tan t - t\)) = t(tan t - t) - log sec t + \(\frac{t^2}2+c\) = t tan t - \(\frac{t^2}2\) - log sec t + c = t tan t - \(\frac{t^2}2\) - log (\(\sqrt{1+tan^2t}+c\)) = x tan-1x - \(\frac12(tan^{-1}x)^2\) - log\(\sqrt {1+x^2}\) + c Hence, \(\int\cfrac{tan^{-1}x}{1+\frac1{x^2}}dx\) = x tan-1x - \(\frac12\) (tan-1x)2 - log\(\sqrt{1+x^2}+c\) |
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