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$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)

factor out the 5, and use partial fraction

$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$

Solve for $A$ and $B$.

$A\left(x-5\right) + B = x$ Then $B-5A$ has to be zero and $A$ has to be 1.

Resulting in

$$ 5 \left[\int \frac{1}{x-5} + \frac{5}{\left(x-5\right)^2}\, \mathrm{d}x \right]$$

$$ \Rightarrow 5 \left[ \ln \vert x - 5 \vert -\frac{5}{x-5}\right] + C$$

However, this approach doesn't give the answer in the book.

Book's Answer

$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right) + C $$

The value should be 30, according to the book.

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up vote 4 down vote accepted

Probably they missed include a constant in book's answer. If we include a constant $k$, the book's answer will change to: $$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k$$ But with some algebra we get $$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k=$$ $$=5(\frac{(x-5)}{x-5}\ln|x-5|-\frac{x}{x-5})+k= 5(\ln|x-5|-\frac{x}{x-5}+1)-5+k=$$ $$=5(\ln|x-5|-\frac{x}{x-5}+\frac{x-5}{x-5})-5+k=5(\ln|x-5|-\frac{5}{x-5})-5+k=$$ $$=5(\ln|x-5|-\frac{5}{x-5})+C$$ Which is your answer, where C is a new constant, such that $C=k-5$.

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Distribute: $$ \frac{5}{x-5}((x-5)\ln|x-5|-x)=5\left(\frac{x-5}{x-5}\ln|x-5|-\frac{x}{x-5}\right). $$ Then $$ \frac{x-5}{x-5}=1. $$

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I don't have the $x$ in the last fraction. – yiyi Nov 29 '12 at 12:20
@MaoYiyi Sorry, didn't notice that. – Joe Johnson 126 Nov 29 '12 at 16:11
@MaoYiyi RicardoCruz has the explanation. – Joe Johnson 126 Nov 29 '12 at 16:14

Your solution is correct, but books solution is also. Differentiate the solutions and you will see, that both of them are Antiderivatives.

Moreover it is: $$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right) = 5 \left(\ln \vert x - 5 \vert - \frac{x-5+5}{x-5}\right) = 5 \left(\ln \vert x - 5 \vert - \frac{5}{x-5}\right) +\mathcal{Const}$$

Optional way to get your solution: $$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x= \frac{5}{2}\int\frac{2x-10}{\left(x-5\right)^2}+\frac{10}{\left(x-5\right)^2}\,\mathrm{d}x=\frac{5}{2}\left(2\ln\vert x-5\vert-\frac{10}{x-5}\right)+\mathcal{C}$$

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how does $x-5 + 5 = 5$? Where did the $x$ goto? – yiyi Nov 30 '12 at 1:02
@MaoYiyi: It is $\frac{x}{x-5} = \frac{x-5+5}{x-5} = 1+\frac{5}{x-5}$ – user127.0.0.1 Nov 30 '12 at 7:41

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