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How do you find the indefinite integral of #int ((x²) / (16-x³)²) dx#?

Answer 1

#(-1/3)*ln(16-x^3)+c#

Don't know if the 2 after the fraction and before the closing bracket is a scalar multiple or an error, so doing it without the 2. replace #16-x^3# with #t# so we have #-3*x^2dx = dt# the equation becomes: #int(-1/(3*t))dt# = #(-1/3)*ln(t)+c# now replacing t back, #(-1/3)*ln(16-x^3)+c#

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Answer 2

Ezra Adams

To find the indefinite integral of (\int \frac{x^2}{(16-x^3)^2} , dx), you can use a substitution method. Let (u = 16 - x^3), then (du = -3x^2 , dx). Solving for (dx), you get (dx = -\frac{du}{3x^2}). Substitute (u) and (dx) into the integral and simplify. You'll end up with an integral in terms of (u), which can be easier to solve. Finally, reverse the substitution to express the result in terms of (x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.