Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀¹/√2 2x arcsin(x²) dx

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Identify the integral to solve: $\int_0^{\frac{1}{\sqrt{2}}} 2x \arcsin(x^2) \, dx$.

Recognize that the integrand is a product of two functions: \$2x$ and $\arcsin(x^2)$. This suggests using integration by parts, where one function is differentiated and the other is integrated.

Choose $u = \arcsin(x^2)$, so that $du = \frac{d}{dx} \arcsin(x^2) \, dx$. Use the chain rule to find $du$: $du = \frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x \, dx = \frac{2x}{\sqrt{1 - x^4}} \, dx$.

Let $dv = 2x \, dx$, then integrate to find $v$: $v = \int 2x \, dx = x^2$.

Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Substitute the expressions for $u$, $v$, and $du$ to rewrite the integral and simplify before evaluating the definite integral from $0$ to $\frac{1}{\sqrt{2}}$.

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