My imagination doesn't help me with $$\int \left(\frac{\arctan x}{\arctan x - x}\right)^3 \mathrm{dx}$$ What tools should I use? W|A doesn't help either.
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2What makes you think that there is an antiderivative for that function? – Ron Gordon May 30 '13 at 21:36
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Hint:
$\because0<\dfrac{\tan^{-1}x}{x}\leq1$ $\forall x\in\mathbb{R}$
$\therefore\int\biggl(\dfrac{\tan^{-1}x}{\tan^{-1}x-x}\biggr)^3~dx$
$=\int\biggl(\dfrac{\dfrac{\tan^{-1}x}{x}}{\dfrac{\tan^{-1}x}{x}-1}\biggr)^3~dx$
$=-\int\dfrac{\dfrac{(\tan^{-1}x)^3}{x^3}}{\left(1-\dfrac{\tan^{-1}x}{x}\right)^3}dx$
$=-\int\dfrac{(\tan^{-1}x)^3}{x^3}\sum\limits_{n=0}^\infty\dfrac{(n+2)(n+1)(\tan^{-1}x)^n}{2x^n}dx$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(n+2)(n+1)(\tan^{-1}x)^{n+3}}{2x^{n+3}}dx$
$=-\int\sum\limits_{n=3}^\infty\dfrac{(n-1)(n-2)(\tan^{-1}x)^n}{2x^n}dx$
doraemonpaul
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