∫_0^1▒x/〖√(x^2+1)〗^2 dx
You asked:
Evaluate the integral: \(\int_{0}^{1}{\frac{x}{{\sqrt{{x}^{2} + 1}}^{2}}}dx\)
MathBot Answer:
Definite integral
\[\int_{0}^{1}{\frac{x}{{\sqrt{{x}^{2} + 1}}^{2}}}dx = \frac{\log{\left(2 \right)}}{2} \approx 0.34657359027997265470861606072909\]