MHB Solving an Integral: Struggling to Make Progress?

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To solve the integral $$\int \frac{x^3 + 1}{x^2 - 4} dx$$, start by performing long division since the numerator's degree is higher than the denominator's. This simplifies the integral to $$\int \left(x + \frac{4x + 1}{x^2 - 4}\right) dx$$. After simplification, the integral can be separated into three parts: $$\int x\,dx$$, $$4\int \frac{x\,dx}{x^2 - 4}$$, and $$\int \frac{dx}{x^2 - 4}$$. Each of these integrals can then be solved individually, leading to the final result.
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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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