Question

Use the table of integrals in this section to find the indefinite integral. $$ \int \frac{2 x}{(1-3 x)^{2}} d x $$

Short answer

The indefinite integral of \( \frac{2 x}{(1-3 x)^{2}} \) is \( \frac{2}{3(1-3x)} + C \).

Step-by-step solution

Identify u and u'

Assign \( u = 1-3x \) and differentiate it to find \( u' \) which gives \( u' = -3 \). Notice that there is a factor of 2 in the numerator of the problem that is not accounted for from \( u' \), we'll address this more in the next step.

Rescaling to match integral form

The integral can now be rewritten as \( -\frac{2}{3} \int \frac{-3x}{(1-3x)^2} dx \) from our identification above. Here, we included \( -\frac{2}{3} \) to match \( u' \) with numerator. Now, the integral is in the form \( \int \frac{u' }{u^{2}} dx \).

Apply the standard integral formula

The standard formula for this type of integral is \( -\frac{1}{u} + C \). Substituting \( u = 1-3x \) back into the equation, we get \( -\frac{1}{1-3x} \) plus the constant of integration \( C \).

Rescale the result

Since we had included the extra factor we now rescale our result to obtain the final answer, i.e. we multiply by \( -\frac{2}{3} \). This results in \( \frac{2}{3(1-3x)}+C \).