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Bit messy but yeah. First you start with a polynomial division, labeled 1, then you write it all out. you then have to use a U-sub for the last term, labeled 2. Finally, you integrate the entire thing and there's your answer.

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U-sub with x^3 -2 as U

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The denominator can be factored down as it is a perfect square. You can then do u-sub on the denominator and solve through integrating.

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complete square inside sqrt and then put it back in and then solve it as sin-1

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since i could not factor the denominator I completed the square. After I used u substitution, integrated, and then plugged x back in for u to get my solution.

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The first step is to use long division since the degree in the numerator is bigger than the denominator. Then, you are able to integrate normally.

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Note that it is nearly in the form of the derivative of arctan, but the x is in the numerator. By doing u-sub it corrects this issue by making the x apart of du.

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Because the degree of the numerator was bigger, I used long division and then took the antiderivative of each of the individual terms I got from dividing.

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Pull the 4 out of the integration. Complete the square on the denominator.  Change the term 16(x-(3/4))^2 into (4x-3)^2. Use (4x-3) for usub and integrate the function as arcsin. Multiply the original 4 by the 4 from the usub to get the solution.

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First you can long divide since the degree is higher on the top then you can separate the terms and integrate seperately.

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complete the square (bottom can't be factored) then sub in for arctangent

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Please post a brief written explanation to explain to your classmates your approach to solving this problem!