LEMNISCATA
Matemàtiques
Femos un cambio de variable:
$$x=t^2\rightarrow dx=2t~dt$$
Sía I la integral a resolver:
$$\displaystyle I=\int\left(\sqrt x\cdot\ln^2x\right)~dx=\int\left(\sqrt{t^2}\cdot\ln^2t^2\right)2t~dt=\int2t^2\ln^2t^2~dt$$
Utilizamos lo metodo d’integración per partes:
$$\begin{array}{lcl}u=\ln^2t^2&\rightarrow&du=2\ln t^2\cdot\dfrac{2t}{t^2}=\dfrac{4\ln t^2}t\\ dv=2t^2~dx&\rightarrow&v=\dfrac{2t^3}3\end{array}$$
$$\displaystyle I=\dfrac{2t^3\ln^2t^2}3-\int\dfrac{2t^3}3\dfrac{4\ln t^2}t~dt=\dfrac{2t^3\ln^2t^2}3-\underbrace{\int\dfrac{8t^2\ln t^2}3~dt}_{I_1}\qquad(1)$$
Utilizamos lo metodo d’integración per partes pa calcular la integral $I_1$:
$$\begin{array}{lcl}u=\ln t^2&\rightarrow&du=\dfrac{2t}{t^2}=\dfrac2t\\ dv=\dfrac{8t^2}3~dt&\rightarrow&v=\dfrac{8t^3}9\end{array}$$
$$\displaystyle I_1=\dfrac{8t^3\ln t^2}9-\int\dfrac{8t^3}9\dfrac2t~dt=\dfrac{8t^3\ln t^2}9-\int\dfrac{16t^2}9~dt~;\\I_1=\dfrac{8t^3\ln t^2}9-\dfrac{16t^3}{27}+k_1$$
Substituyindo en (1):
$$I=\dfrac{2t^3\ln^2t^2}3-\left(\dfrac{8t^3\ln t^2}9-\dfrac{16t^3}{27}+k_1\right)~;\\I=\dfrac{18t^3\ln^2t^2-24t^3\ln t^2+16t^3}{27}+k~;\\I=2t^3\cdot\dfrac{9\ln^2t^2-12\ln t^2+8}{27}+k$$
Sabendo que $x=t^2$, tenemos que $t^3=x\sqrt x$. Desfemos lo cambio de variable:
$$\boxed{I=2x\sqrt x\cdot\dfrac{9\ln^2x-12\ln x+8}{27}+k}$$