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Perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral.

∫ 3y⁴ +9y²-1 /(y³ +3y) dy

User Arul Suju
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Final answer:

To evaluate the integral ∫ (3y⁴ + 9y² - 1) / (y³ + 3y) dy, perform long division, write the fraction as a sum of partial fractions, and then evaluate the integral.

Step-by-step explanation:

To evaluate the integral ∫ (3y⁴ + 9y² - 1) / (y³ + 3y) dy, we first need to perform long division on the integrand. Dividing y³ + 3y into 3y⁴ + 9y² - 1 gives 3y. We can rewrite the integrand as 3y + 27 / (y³ + 3y).

Next, we write the proper fraction as a sum of partial fractions. Since the denominator is a cubic polynomial, we can express it as a sum of linear factors: y³ + 3y = y(y² + 3). The partial fraction decomposition becomes:

3y + 27 / (y³ + 3y) = A/y + (By + C)/(y² + 3)

Finally, we can evaluate the integral using the partial fraction decomposition. The integral of A/y with respect to y is A ln|y|, and the integral of (By + C)/(y² + 3) can be expressed in terms of inverse trigonometric functions.

User Shokha
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