RUStudent said:
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The integral of (32/((x^2)(sqrt(16-(x^2)))))dx is equal to 16arcsin(x/4) + C, where C is the constant of integration.
To solve this integral, you can use the substitution method by letting u = x/4 and du = dx/(4sqrt(16-(x^2))). Then, the integral becomes 8*int(1/(u^2)*du), which can be easily solved using the power rule.
No, the integral is not defined for all values of x. It is only defined for values of x such that 16-(x^2) > 0, which means that x must be between -4 and 4.
Yes, the indefinite integral of (32/((x^2)(sqrt(16-(x^2)))))dx is equal to 16arcsin(x/4) + C, where C is the constant of integration.
The integral represents the area under the curve of the given function. It can also be used to find the antiderivative of the function, which is useful in solving differential equations and other mathematical problems.