Find the value of $m$ given that $\displaystyle\int_0^m \dfrac{dx}{3x+1}=1$

Question

I have to find the value of $m$ such that:

$\displaystyle\int_0^m \dfrac{dx}{3x+1}=1.$

I'm not sure how to integrate when dx is in the numerator. What do I do?

edit: I believe there was a typo in the question. Solved now, thank you!

Answer 1

That was probably a typo (meaning the double $\mathrm{d}x$) and your integral is the same as the following:

$$ I = \int^m_0 \frac{1}{3x+1} \, \mathrm{d}x = \int^m_0 \frac{\mathrm{d}x}{3x+1} = \int^m_0 \mathrm{d}x \, \frac{1}{3x+1} $$

You can place $\mathrm{d}x$ in the numerator, or even before the fraction while there's no confusion (the latter is less common). It's just a matter of notation and convention.

Can you take it from here?

Answer 2

It's like integrating $\int f(x)dx$ with $f(x) = \frac{1}{3x+1}$

Answer 3

Since: $$\int_{0}^{m}\frac{dx}{3x+1}=\frac{1}{3}\log(3m+1)$$ we have: $$ m = \frac{e^3-1}{3}.$$