How do you integrate #((x^2) / (16-x^3)^2) dx#?

Answer 1

#1/(3(16-x^3))+C#

Choose #t=16-x^3#, then #dt=-3x^2dx => x^2dx=-dt/3# #intx^2/(16-x^3)^2dx=int1/t^2(-dt/3)=-1/3intt^(-2)dt=# #=-1/3 t^(-1)/-1+C=1/3 1/t+C=1/(3(16-x^3))+C#
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Answer 2

To integrate ((x^2) / (16-x^3)^2) dx, you can use a substitution method. Let u = 16 - x^3. Then, differentiate both sides with respect to x to find du/dx. After finding du/dx, solve for dx in terms of du. Next, substitute u and du into the integral and integrate with respect to u. Finally, revert back to the variable x by substituting u = 16 - x^3 back into the result obtained from integrating with respect to u.

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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.

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.

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.

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.

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