# How do you integrate #int-(8x^3)/(-2x^4+5)^5# using substitution?

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We have:

When choosing a good substitution for functions such as these, which have mainly terms that resemble polynomial functions (i.e., not trigonometric or exponential), it's good to find powers that are one apart, since the lower can often be the derivative of the higher.

Also, its good to write a fraction's denominator, if it's multiple terms, as a single term via substitution. Both of these pieces of advice apply here:

The substitution can now take place easily:

This can be solved using typical integration rules:

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To integrate the function (\int \frac{-8x^3}{(-2x^4 + 5)^5}) using substitution, we can let (u = -2x^4 + 5). Then, (du = -8x^3 dx). Rearranging for (dx), we get (dx = \frac{du}{-8x^3}).

Now, substituting (u) and (dx) into the integral, we have:

(\int \frac{-8x^3}{(-2x^4 + 5)^5} dx = \int \frac{1}{u^5} du).

Integrating (\frac{1}{u^5}) with respect to (u), we get:

(\int \frac{1}{u^5} du = \frac{1}{-4u^4} + C).

Now, substituting back for (u = -2x^4 + 5), we get:

(\frac{1}{-4(-2x^4 + 5)^4} + C).

Thus, the integral of (\int \frac{-8x^3}{(-2x^4 + 5)^5}) using substitution is (\frac{1}{-4(-2x^4 + 5)^4} + C), where (C) is the constant of integration.

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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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