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The problem is:


I got and am fairly confident in:

$$\frac74 \ln|2x+5| + \frac{35}{8x + 20}$$

However this is apparently not the correct answer. Im not being graded on this and it is not a homework problem for me. Does anyone get a different answer than I did or is the key incorrect?

Thanks for the responses. I'm convinced my friend entered it into the computer wrong. According to her the constant was supposed to be ommitted, so either she entered it wrong or her professor did.

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It's correct, aside from not having a constant of integration. You can check this: take the derivative of your expression and show you obtain $7x\over (2x+5)^2$. – David Mitra Sep 19 '12 at 4:33
One can imagine the log term to be called $\frac{7}{4}\ln(|x+5/2|)$, which would differ by a constant from your version. – André Nicolas Sep 19 '12 at 5:31
up vote 2 down vote accepted

First let’s dispose of a couple of notational problems: the integral should be $$\int\frac{7x}{(2x+5)^2}dx\;,$$ and you’re missing a constant of integration in your answer.

This is not a partial fractions problem, though it may have arisen from one: the integrand is already decomposed as far as it can be. Just substitute $u=2x+5$, so that $du=2dx$. Then $x=\frac12(u-5)$, and $dx=\frac12du$, so your integral becomes

$$\begin{align*} \int\frac{\frac72(u-5)}{u^2}\frac12du&=\frac74\int\frac{u-5}{u^2}du\\ &=\frac74\int(u^{-1}-5u^{-2})du\\ &=\frac74\left(\ln|u|+5u^{-1}\right)+C\\ &=\frac74\left(\ln|2x+5|+\frac5{2x+5}\right)+C\\ &=\frac74\ln|2x+5|+\frac{35}{4(2x+5)}+C \end{align*}$$

In other words, yes, apart from the missing constant of integration your answer is correct.

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I simplified the denominator in my post – rmh52 Sep 19 '12 at 4:43

Your answer seems to be correct. +C?

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