Find the indefinite integral.∫3−y22ydy\int_{}^{}\frac{3-y^2}{2y}dy∫2y3−y2dy

32ln⁡∣y∣−14y2+C\frac32\ln\left|y\right|-\$$frac14y^2+C23$$ln∣y∣−41y2+C

Find the indefinite integral. ∫ 3 − y 2 2 y d y \int_{}^{}\frac{3-y^2}{2y}dy ∫ 2 y 3 − y 2 d y

Here we're asked to find the indefinite integral of three minus y squared over two y d y. Now on first glance, it looks like a rather complicated integral that has a lot going on specifically because we're dealing with a fraction. Now we can actually break this fraction down and make this a bit easier. Because we have two terms in our numerator and just one term in the denominator, we can split this up into two separate fractions, making this the integral of three over two Y minus y squared over two y. Then we're going to see some simplification happen because one of these y squareds is going to cancel out that y on the bottom. So this is really three halves times one over y just showing that constant out there by itself and then minus one half y just again breaking that out into a constant and whatever variable we're working with, then integrating with respect to y. So this looks like a much easier integral to deal with and we can now go ahead and apply our rules for integration. So for this first one, three halves is just a constant, so that's gonna stay the same. Then as I find the antiderivative of one over y, I can apply the rule that we just learned to get here the natural log of the absolute value of y. Then here using the power rule, keeping that constant out front, one half, then applying the power rule to this y, adding one to that exponent, then dividing by that new exponent, that gives us one half y squared. Then we wanna add on our constant of integration c, and we can do a bit of simplification here because one half times one half is really just one fourth. And now we have our final answer, three halves natural log of the absolute value of y minus one fourth y squared plus c. Let us know if you have any questions and let's keep going.

Find the indefinite integral.∫3−y22ydy\int_{}^{}\frac{$$3-y^2$$}{2y}dy∫2y3−y2dy

32ln⁡∣y∣−14y2+C\frac32\ln\left|y\right|-\$$frac14y^2+C23$$ln∣y∣−41y2+C

32+14y2+C\frac32+\frac{1}{$$4y^2$$}+C23+4y21+C

32−14y2+C\frac32-\frac{1}{$$4y^2$$}+C23−4y21+C

32ln⁡∣y∣−14y2+C\frac32\ln\left|y\right|-\frac{1}{$$4y^2$$}+C23ln∣y∣−4y21+C

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Business Calculus

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

Find the indefinite integral.
$$\int$_{}^{}$\frac{3-y^2}{2y}$dy$

$$\frac$32$\ln\]\left$|y$\right$|-$\frac$14y^2+C$

$$\frac$32+$\frac{1}{4y^2}$+C$

$$\frac$32-$\frac{1}{4y^2}$+C$

$$\frac$32$\ln\]\left$|y$\right$|-$\frac{1}{4y^2}$+C$

Verified step by step guidance

Step 1: Begin by rewriting the given integral for clarity. The integral is ∫(3 - y^2) / (2y) dy. Split the fraction into two separate terms: ∫(3 / (2y)) dy - ∫(y^2 / (2y)) dy.

Step 2: Simplify each term. For the first term, 3 / (2y) simplifies to (3/2)(1/y). For the second term, y^2 / (2y) simplifies to y / 2. The integral now becomes ∫(3/2)(1/y) dy - ∫(y/2) dy.

Step 3: Integrate each term separately. For the first term, ∫(3/2)(1/y) dy, use the rule ∫(1/y) dy = ln|y|. This gives (3/2)ln|y|. For the second term, ∫(y/2) dy, use the power rule ∫y^n dy = y^(n+1)/(n+1). Here, n = 1, so the result is (1/2)(y^2/2) = y^2/4.

Step 4: Combine the results of the two integrals. The integral becomes (3/2)ln|y| - (1/4)y^2 + C, where C is the constant of integration.

Step 5: Verify the solution by differentiating the result. Differentiate (3/2)ln|y| - (1/4)y^2 + C to ensure it matches the original integrand (3 - y^2) / (2y). This confirms the solution is correct.

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Find the indefinite integral.∫3−y22ydy\(\int\)_{}^{}\(\frac{3-y^2}{2y}\)dy∫2y3−y2dy

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Find the indefinite integral.
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