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How to integrate x^3/(x^2+4)?

To integrate x^3/(x^2+4), use substitution with u=x^2+4 and du=2xdx.

To integrate x^3/(x^2+4), we can use substitution. Let u=x^2+4, then du/dx=2x. Rearranging, we get dx=du/2x. Substituting these into the integral, we get:

∫x^3/(x^2+4) dx = ∫(u-4)/(2u) du

We can split this into two integrals:

∫(u-4)/(2u) du = ∫u/(2u) du - ∫4/(2u) du

Simplifying, we get:

∫(u-4)/(2u) du = 1/2 ∫1 du - 2 ∫1/u du

Integrating, we get:

∫(u-4)/(2u) du = 1/2 u - 2 ln|u| + C

Substituting back in for u, we get:

∫x^3/(x^2+4) dx = 1/2 (x^2+4) - 2 ln|x^2+4| + C

Therefore, the integral of x^3/(x^2+4) is 1/2 (x^2+4) - 2 ln|x^2+4| + C.

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