Evaluar $\int_0^R \frac{r^2}{(1+r^2)^2}dr.$

Question

Estoy tratando de evaluar la siguiente integral: $$\int_0^R \frac{r^2}{(1+r^2)^2}dr.$$ I might substitute $u=r^2$, but I don't find $du$ anywhere. Obviously the integral should be bounded on $R\[0,\infty)$.

Alguna idea?

Answer 1

Sustituto $r=\tan{t}$; $dr = \sec^2{t} \, dt$. Entonces la integral es igual a

$$\int_0^{\arctan{R}} dt \, \sec^2{t} \, \frac{\tan^2{t}}{ \sec^4{t}} = \int_0^{\arctan{R}} dt \,\sin^2{t}$$

que luego es sencillo para evaluar:

$$\frac12 \left [t - \sin{t} \cos{t} \right ]_{0}^{\arctan{R}} = \frac12 \left [\arctan{R} - \frac{R}{1+R^2}\right ]$$

Answer 2

Denominador contiene $1+r^2,$ pongámonos $r=\tan x$. Entonces $$\begin{multline} \int_0^R\frac{r^2}{(1+r^2)^2}\,dr =\int_0^{\arctan R}\frac{\tan^2x}{\sec^4x}\sec^2x\,dx =\int_0^{\arctan R}\sin^2x\,dx\\ =\int_0^{\arctan R}\frac{(1-\cos2x)}2\,dx =\frac12\left(x-\frac {\sin2x}2\right)_0^{\arctan R} \\ =\frac12\left(\arctan R-\frac R{1+R^2} \right) \end{multline}$$ como $$\sin(\arctan R)=2\frac{\tan(\arctan R)}{1+\tan^2(\arctan R)}=\frac{2R}{1+R^2}$$

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Alternativamente, integrando por partes, $$\begin{multline} \int\frac{r^2}{(1+r^2)^2}\,dr =\int r\cdot \frac r{(1+r^2)^2}\,dr \\ =r\cdot \frac{-1}{2(1+r^2)}-\int \left(\frac{dr}{dr}\cdot \frac{-1}{2(1+r^2)}\right)\,dx \end{multline}$$

$$=\frac{-r}{2(1+r^2)}+\frac12\int\frac{dr}{1+r^2}=-\frac r{2(1+r^2)}+\frac12\arctan r$$

Answer 3

Sugerencia:$$\int_0^R \frac{r^2}{(1+r^2)^2}dr=\int_0^R \frac{r^2+1-1}{(1+r^2)^2}dr=\color{red}{\int_0^R \frac{1}{(1+r^2)}dr}-\color{green}{\int_0^R \frac{1}{(1+r^2)^2}dr}$$ respuesta de la red integral es obvio .

para el verde integral sustituto $r=\sin hx\to dr=\cos hx$ $$\color{green}{\int_0^R \frac{1}{(1+r^2)^2}dr=\int_0^R \frac{\ coshx}{(1+\sinh x^2)^2}=\int_0^R \frac{dx}{cos^3hx}=\int_0^R \frac{dx}{\ coshxcos^2hx}=\int_0^R\ sechx(1-\tan^2h x)}=....... $$