Răspuns:
Varianta d. 512.
Explicație pas cu pas:
Fie [tex]I=\int_0^{1024}\dfrac{\ln(2017-x)}{\ln[1505^2-(512-x)^2]}dx[/tex].
Cu substitutia [tex]t=1024-x,\:\:dx=-dt[/tex], limitele se schimba in [tex]1024-1024=0[/tex] si [tex]1024-0=1024[/tex], deci:
[tex]I=\int_0^{1024}\dfrac{\ln(2017-x)}{\ln[1505^2-(512-x)^2]}dx\\\\I=\int_{1024}^0\dfrac{\ln(2017-(1024-t))}{\ln\{1505^2-[512-(1024-t)]^2\}}(-1)dt\\\\I=\int_{0}^{1024}\dfrac{\ln(2017-(1024-t))}{\ln\{1505^2-[512-(1024-t)]^2\}}dt\\\\I=\int_{0}^{1024}\dfrac{\ln(993+t)}{\ln[1505^2-(512-t)^2]}dt\\\\\text{Dar stim din ipoteza ca } I=\int_{0}^{1024}\dfrac{\ln(2017-t)}{\ln[1505^2-(512-t)^2]}dt\\\\\\\text{Asadar } 2I=I+I=\int_{0}^{1024}\dfrac{\ln(2017-t)+\ln(993+t)}{\ln[1505^2-(512-t)^2]}dt[/tex]
[tex]2I=\int_{0}^{1024}\dfrac{\ln[(2017-t)(993+t)]}{\ln[(1505-512+t)(1505+512-t)]}dt\\\\2I=\int_{0}^{1024}\dfrac{\ln[(2017-t)(993+t)]}{\ln[(993+t)(2017-t)]}dt=\int_{0}^{1024}dt=1024\\\\I=512[/tex]
