Integrating for a solution in terms of an natural logarithm

Question

Evaluating the following integral:
$$\int_1^2 \frac2{1-3x}\ dx$$

why do you have to take the factor of $-2/3$ out when evaluating the integral?

Answer 1

Suppose you didn't take that factor out, you would be tempted to try: $$\int_1^2 \frac 2{1-3x} dx = -\frac{2}{3}\ln(1-3x)\bigg|_1^2 = -\frac 23\left[\ln(-5) - \ln(-1)\right]$$ But this can't be... because $\ln(u)$ is undefined for $u\leq 0$.

If you remove this factor: $$\int_1^2 \frac 2{1-3x}\ dx = -\frac 23\int_1^2 \frac{1}{x - \frac13}\ dx = -\frac 23 \left[\ln\left(\frac 53\right) - \ln\left(\frac 23\right)\right] = \frac 23 \ln\left(\frac{2}{5}\right)$$

Answer 2

Consider we do a $u$ substitution.

$u=1-3x$

$du=-3dx$

We have $2dx$ but not $-3dx$

How do we get the $-3dx$?

Notice $(\frac{-2}{3})(-3)dx=2dx$

Answer 3

Technically, you do not have to factor the $-2/3$ out of the equation. You could still leave in $-2/3$ and integrate, which would cause you to divide by $-3$ which would give you the same answer.

Answer 4

Using the substitution method, let $u=1-3x$. Then differentiate $u$ with respect to $x$ to get $dx = -\frac{1}{3} du$. Substitute away $dx$ and $1-3x$ in your integral to obtain $$-\frac{2}{3} \int_{1}^{2} \frac{1}{u}\,du.$$

Then replace your upper and lower limits of $x$ with that of $u$ to obtain $$-\frac{2}{3} \int_{-2}^{-5} \frac{1}{u}\,du,$$

and then take the integral of $u$.

So as you can see that if you do not take $-\frac{2}{3}$ out of the integral then you cannot apply the substitution method to evaluate the integral.