Solving an Integral: Struggling to Make Progress?

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SUMMARY

The integral $$\int \dfrac{x^3+1}{x^2-4} dx$$ can be solved effectively by first performing polynomial long division due to the numerator's higher degree. After dividing, the integral simplifies to $$\int \left(x + \frac{4x+1}{x^2-4}\right)dx$$. This allows for further integration using partial fractions or trigonometric substitution. The denominator can be factored as $(x + 2)(x - 2)$, which aids in the integration process.

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I have this integral:

$$\frac {x^3 + 1}{x^2 - 4}$$

And I'm unsure how to approach it. I can factor the denominator like this $(x + 2)(x - 2)$ but I'm not sure if this is useful.
 
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$$\int \dfrac{x^3+1}{x^2-4} dx$$

Since the numerator has higher degree than the denominator, just do a long division. Once you've divided, you can do partial fractions or a trig substitution.
 
tmt said:
I have this integral:

$$\frac {x^3 + 1}{x^2 - 4}$$

And I'm unsure how to approach it.
I can factor the denominator like this $(x + 2)(x - 2)$
but I'm not sure if this is useful.
Use long divison and the integral becomes:. . \int \left(x + \frac{4x+1}{x^2-4}\right)dx

Then you can integrate these three integrals: .\int x\,dx + 4\int\frac{x\, dx}{x^2-4} + \int\frac{dx}{x^2-4}

 

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