Solving $\int_0^1 \int_0^x x \sqrt{x^2+3y^2} \,dy\, dx $

Note that $$ \int_0^1 \int_{0}^{x} x \sqrt{x^2+3y^2} \,dy\, dx = \int_0^1 x \left(\int_{0}^{x} \sqrt{x^2+3y^2} \ dy\right) dx $$ and the inner integral can be taken using identity (29) at http://integral-table.com: $$ \int \sqrt{a^2+y^2} \ dy = \frac{y}{2} \sqrt{a^2+y^2} + \frac{a^2}{2} \ln \left| y + \sqrt{y^2 + a^2} \right| + C $$

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Integration

Multivariable Calculus

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