Question

[23 pct] Va rog frumos. Sa inteleg si cum ati facut.

Answer 1

Răspuns:

$\frac{1}{2}ln(1+\sqrt2)$

Explicație pas cu pas:

$I=\int\limits^1_0 {\frac{x}{\sqrt{x^4+1}} \, dx$

Notam: $x^2=t~=>~2xdx=dt~=>~xdx=\frac{1}{2}dt$.

Daca x=0 => t=0.

Daca x=1 => t=1.

Integrala devine:

$I=\frac{1}{2}\int\limits^1_0 {\frac{1}{\sqrt{t^2+1}} \, dt=\frac{1}{2}ln(t+\sqrt{t^2+1})|^1_0=\frac{1}{2}[ln(1+\sqrt2)-ln(0+1)]=\frac{1}{2}ln(1+\sqrt2)$