# Evaluating $\int \left(\frac{3}{5} - \frac{8}{x}\right) \ dx$

I am unsure how to get the antiderivative of

$$f(x) = \frac3 5 − \frac8 x$$

I know the answer is $-8\ln(|x|) + C + \frac{3x}{5}$ as it says in my textbook.

But I am unsure how I get something like $\ln(x)$ in the answer? Could someone guide me through this? I am totally lost as for other antiderivatives I have no problem but this stuns me.

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Is it because the 8 is over an x? – soniccool Jul 11 '12 at 0:34

First: there is not such thing as "the antiderivative of $f(x)$". There are generally infinitely many different functions that are antiderivatives of $f(x)$. What you can talk about is either "the family of antiderivatives" or "the most general antiderivative", or else "an antiderivative". It seems clear you are looking for "the most general antiderivative."

Remember that the derivative of $\ln(x)$ is $\frac{1}{x}$. And the derivative of $kf(x)$ with $k$ a constant is $kf'(x)$.

So to get a derivative equal to $-\frac{8}{x} = -8\left(\frac{1}{x}\right)$, you can take $-8f(x)$, where $f(x)$ is a function whose derivative is $\frac{1}{x}$. For example.... $\ln(x)$.

(In fact, the most general function whose derivative is $\frac{1}{x}$ is $\ln|x|+C$, because $(\ln(-x))' = \frac{1}{-x}(-x)' = \frac{1}{x}$, using the Chain Rule.)

So we can, and should, take $-8\ln|x|$ to get the $-\frac{8}{x}$ part of the most general antiderivative. For $\frac{3}{5}$, we can take $\frac{3}{5}x$. And finally, a sum of antiderivatives is an antiderivative of a sum because the derivative of a sum is the sum of antiderivatives; so we take the sum of the two parts, and add the constant to get the general antiderivative. We have: $$\underbrace{\frac{3}{5}x}_{\text{antiderivative of }\frac{3}{4}} -\underbrace{8\ln|x|}_{\text{antiderivative of }\frac{8}{x}} + C.$$

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Oh alright i get it because $1/x$ would also be as ln(x) – soniccool Jul 11 '12 at 0:46
The most general such function is not $\ln|x|+C$ but (if we're dealing only with real variables) $\ln|x|+\begin{cases} A & \text{if }x>0,\\ B & \text{if }x<0.\end{cases}$ – Michael Hardy Jul 14 '12 at 2:29

You need to use the fact the derivative is linear $$\left( \color{orange}{k}\cdot \color{purple}{f(x)}+\color{orange}{j}\cdot \color{blue}{g(x)}\right)'=\color{orange}{k}\cdot\color{purple}{f'(x)}+\color{orange}{j}\cdot\color{blue}{g'(x)}$$

You seem to know that

$$(x)'=1$$

$$(\log x)'=\dfrac 1 x$$

Then, using the above, can you find a primitive of

$$f(x)=\frac 3 5 \cdot \color{red}{1}+(-8)\cdot \color{red}{\frac 1 x}\text{ ? }$$

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The antiderivative of $\dfrac{1}{x}$ is $\ln|x|$. You can break up the antiderivatives by addition, subtraction, or you can pull scalar multiples out to the front. Since $-\dfrac{8}{x}$ is just $-8 \cdot \dfrac{1}{x}$, the antiderivative is $-8 \ln|x| + C$.

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