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Problem 35 Hard Difficulty

Verify Formula 53 in the Table of Integrals (a) by differentiation and (b) by using the substitution $ t = a + bu $.

Answer

(a) $\frac{u^{2}}{(a+b u)^{2}}$
(b) $\frac{1}{b^{3}}\left(a+b u-\frac{a^{2}}{a+b u}-2 a \ln |a+b u|\right)+C$

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Video Transcript

Okay, So this question wants us to derive the integration formula on the table for this form. So it says to make this substitution t equals a plus b U So from there we see the d t is equal to be Do you or DT over B is equal to do you. And then we also might need to know what you is. So if we solve for you in the top equation, we get t minus a all over B. So now we can rewrite are integral in the T world, so we'll get ah, you squared on top, which is t minus a over B squared over t squared DT with a factor of one over b. So now we can expand this in a girl out. So we get a one over, be cubed, We can pull the b squared out from our numerator here. So we have the integral both t minus a quantity squared over T square D t. And this turns into one over be cubed times theater girl of T squared minus to a T plus a squared all over T squared DT. And now we'll just divide each term by t squared into, so we'll have three different intervals. So we have won over B cubed times. The integral of T squared over T squared is one plus. The integral of T divided by t squared is So you're negative to a over tea then plus the integral of a squared over t squared You too. And we know how to evaluate each of these separately. So we get one over, be cubed out in front times T minus two, a Ellen of tea plus negative a squared over TV plus c. So now let's back substitute in remembering that tea is equal to a plus b u. And this is exactly the formula and our integration table. And another way to check it is to take the derivative of this and show that D D you of f of you is equal to lower case off of you. And in this case, f of you is equal to you squared over a plus b u quantity squared. So you can also show that just remember, they'll have to use the chain role for a lot of these

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