$$\int_{1}^{10}(2624 / (x^2 +x+1)) dx$$

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on Oct 25, 2024 at 5:06 pm



You asked:

Evaluate the integral: \(\int_{1}^{10}{\frac{2624}{{x}^{2} + x + 1}}dx\)

MathBot Answer:

Definite integral

\[\int_{1}^{10}{\frac{2624}{{x}^{2} + x + 1}}dx = - \frac{5248 \pi \sqrt{3}}{9} + \frac{5248 \sqrt{3} \arctan{\left(7 \sqrt{3} \right)}}{3} \approx 1337.12945768727539028547237373335125\]

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