Całka nieoznaczona.

[szymonzbir]

Witam, proszę o pomoc z następującą całką.
\[\int \left(\frac{x}{\arctan(x)}-1\right)^{-2}dx\]

[janusz55]

Podstawienia:

\( x = \tg(t), \ \ dx = \frac{1}{\cos^2(t)} dt, \ \ t\in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right). \)

[janusz55]

\( \int\frac{1}{\left(\frac{\tg(t)}{t} -1\right)^2 \cos^2(t)} dt \)

\( \tg(t) = u, \ \ -\frac{\pi}{2} < u < \frac{\pi}{2}.\)

[janusz55]

\( \int\frac{1}{\left(\frac{\tg(t)}{t} -1\right)^2 \cos^2(t)} dt \)

\( \tg(t) = u, \ \ -\frac{\pi}{2} < t < \frac{\pi}{2}.\)

[szymonzbir]

Ale podstawienie za tg(t)=u spowoduje, że wrócimy do tej samej całki, nic się nie uprości.

[Icanseepeace]

\( \int \left( \frac{\arctan(x)}{x - \arctan(x)} \right)^2 dx = \int \left(1 - \frac{x}{x - \arctan(x)} \right)^2 dx = \\
\int 1 dx - \int \left( \frac{2x}{x - \arctan(x)} \right) + \int \left(\frac{x}{x - \arctan(x)} \right)^2 dx \stackrel{*}{=} \\
\int 1 dx -\biggl( \frac{x^2 + 1}{x - \arctan(x)} +\int \left(\frac{x}{x - \arctan(x)} \right)^2 \biggl) + \int \left(\frac{x}{x - \arctan(x)} \right)^2 dx = \\
x - \frac{x^2 + 1}{x - \arctan(x)} + C
\)
Równość oznaczona przez \( * \) wynika z wzoru
\( \int \frac{dv}{u} = \frac{v}{u} + \int \frac{v}{u^2}du \)
dla \( v = x^2 + 1 \) oraz \( u = x - \arctan(x) \)

[janusz55]

\( \int \frac{1}{\left(\frac{x}{\arctg(x)} -1\right)^2} dx = \int \frac{arctg^2(x)}{ x^2 -2x\cdot arctg(x) +arctg^2(x)} = \int \frac{\arctg^2(x)}{[x -\arctg(x)]^2} dx \)

\( \arctg(x) = t, \ \ x = tg(t),\ \ dx = \frac{1}{\cos^2(t)}dt. \)

\( \int\frac{t^2}{[t -\tg(t)]^2} \cdot \frac{1}{\cos^2(t)} dt = \int \frac{t^2}{\left[t - \frac{\sin(t)}{\cos(t)}\right]^2\cos^2(t)} dt =\int \frac{t^2}{\left[t^2 -2t\frac{\sin(t)}{\cos(t)}+ \frac{\sin^2(t)}{\cos^2(t)}\right]\cos^2(t)} dt = \int\frac{t^2}{t^2\cos^2(t) -2t\sin(t)\cos(t) +\sin^2(t)} dt = \int\frac{t^2}{[t\cos(t)-sin(t)]^2} dt\)

\( t^2 = t^2\sin^2(t) + t^2\cos^2(t),\ \ 0 = -t\sin(t)\cos(t) + t\sin(t)\cos(t)\)

\( \int \frac{-t\sin(t)\cos(t) + t^2\cos^2(t) + t^2\sin^2(t) + t\sin(t)\cos(t) }{(\sin(t) -t\cos(t))^2} dt = \frac{-t\cos(t)[\sin(t)-t\cos(t)] -t\sin(t)[-t\sin(t)-\cos(t)]}{(\sin(t) -t\cos(t))^2}dt = \int \left[ -\frac{t\cos(t)}{\sin(t)-t\cos(t)} - \frac{t\sin(t)(-t\sin(t)-\cos(t))}{(\sin(t)-t\cos(t))^2} \right]dt = \)

\( = \int \frac{t\sin(t)(t\sin(t)+\cos(t))}{(\sin(t)-t\cos(t))^2} dt - \int \frac{t\cos(t)}{\sin(t)-t\cos(t)} dt \)

Pierwszą z całek obliczamy metodą całkowania przez części

\( f(t) = t \sin(t) +\cos(t), \ \ g'(t) = \frac{t\sin(t)}{[\sin(t)+t\cos(t)] ^2} \)

\( f'(t) = t \cos(t), \ \ g(t) = -\frac{1}{\sin(t) -t\cos(t)} \)

\(\int \frac{t\sin(t)(t\sin(t)+\cos(t))}{(\sin(t)-t\cos(t))^2} dt = -\frac{t\sin(t) +\cos(t)}{\sin(t)-t\cos(t)} - \int -\frac{-t\cos(t)}{\sin(t)-t\cos(t)}dt - \int\frac{t\cos(t)}{|sin(t)-t\cos(t)} = -\frac{t\sin(t) +\cos(t)}{\sin(t)-t\cos(t)}.\)

Stąd

\(\int\frac{t^2}{(t\cos(t)-sin(t))^2} dt = -\frac{t\sin(t) +\cos(t)}{\sin(t)-t\cos(t)} + c. \)


\( \int\frac{t^2}{[1-\tg(t)]^2}\cdot \frac{1}{\cos^2(t)} dt = \frac{t\sin(t) +\cos(t)}{t\cos(t) - sin(t)} +c. \)

Wracamy do całki wyjściowej z funkcją podcałkową argumentu \( x \) i arkusem tangensem tego argumentu

\( \int\frac{dx}{\left(\frac{x}{\arctg(x)} -1\right)^2} = \frac{\arctg(x) \cdot \sin(\arctg(x))+ \cos(\arctg(x))}{\arctg(x)\cos(\arctg(x))-\sin(\arctg(x))} + c. \)

Uwzględniamy równości:

\( \sin(\arctg(x)) = \frac{x}{\sqrt{1+x^2}} , \ \ \cos(\arctg(x)) = \frac{1}{\sqrt{1+x^2}}.\)

\( \int\frac{dx}{\left(\frac{x}{\arctg(x)} -1\right)^2} = \frac{x\frac{\arctg(x)}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}}{\frac{\arctg(x)}{\sqrt{1+x^2}} - \frac{x}{\sqrt{1+x^2}}} + c = \frac{x\arctg(x) +1}{\arctg(x) -x} + c . \)