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This is the problem:

$$ \int \frac{dx}{ax+b}$$ where $$a \ne 0$$

So maybe...

$u = ax + b$ so $\frac{du}{dx} = a$ so $du = a \cdot dx$ so $ \frac{du}{a} = dx$

$$ = \frac{1}{a} \int \frac{1}{u} \cdot du$$

$$= \frac{1}{a} \cdot ln|u| $$

$$= \frac{1}{a} ln |ax+b| + C$$

Does that look right?

Why does $ln|u| = \frac{1}{x}$? Why isn't it just $ln(u)$?

Jwan622
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  • Are you asking why it is the absolute value of $|u|$ and not $u$? If it is your question, then because you did not determine how $a$ and $b$ change with respect to x. And it is absolute value to make sure Ln argument is always positive. – Fatimah Apr 14 '18 at 16:18

3 Answers3

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Note that $$ \frac{d}{dx}\left( \frac{1}{a}\ln|ax+b| \right)=\frac{1}{a}\frac{1}{ax+b}\frac{d}{dx}(ax+b)=\frac{a}{a}\frac{1}{ax+b}=\frac{1}{ax+b} $$ so you are correct.

Sri-Amirthan Theivendran
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I suppose that your problem is with the absolute value sign.

If $x>0$, $|x|=x$ and hence

$$\frac{d\ln|x|}{dx}=\frac{d\ln x}{dx}=\frac{1}{x}$$

If $x<0$, then $-x>0$ and $|x|=-x$.

$$\frac{d\ln|x|}{dx}=\frac{d\ln (-x)}{d(-x)}\cdot\frac{d(-x)}{dx}=\frac{1}{-x}\cdot(-1)=\frac{1}{x}$$

Therefore, $\displaystyle \int\frac{1}{x}dx=\ln|x|+C$.

If $x>0$, we still have $\displaystyle \int\frac{1}{x}dx=\ln x+C$.

CY Aries
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Intuitively:

Note that $\frac{1}{u}$ is defined for $u\ne 0$, but $\log (u)$ is defined only for $u\ge 0$ so, if we want the primitive of the function for all values where it is defined we need $log |u|$.

Emilio Novati
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