∫_0^1▒x/〖√(x^2+1)〗^2 dx

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on Jan 15, 2025 at 1:48 pm



You asked:

Evaluate the integral: ∫01xx2+12dx\int_{0}^{1}{\frac{x}{{\sqrt{{x}^{2} + 1}}^{2}}}dx

MathBot Answer:

Definite integral

∫01xx2+12dx=log⁡(2)2≈0.34657359027997265470861606072909\int_{0}^{1}{\frac{x}{{\sqrt{{x}^{2} + 1}}^{2}}}dx = \frac{\log{\left(2 \right)}}{2} \approx 0.34657359027997265470861606072909

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