1.f(x)=∫0xg(t)loge(1−t1+t)dtf(x)=\int_0^x g(t)\log_e(\frac{1-t}{1+t})dtf(x)=∫0xg(t)loge(1+t1−t)dt, ggg continuous odd. If ∫−π/2π/2(f(x)+x2cosx1+ex)dx=(πα)2−α\int_{-\pi/2}^{\pi/2}(f(x)+\frac{x^2\cos x}{1+e^x})dx = (\frac{\pi}{\alpha})^2-\alpha∫−π/2π/2(f(x)+1+exx2cosx)dx=(απ)2−α, then α\alphaα isa.111b.222c.333d.444Login to continueOnly logged in users canattempt or see the solution.
