# How do you use Integration by Substitution to find intdx/(5-3x)?

Jul 25, 2014

$- \frac{1}{3} \cdot \ln \left(5 - 3 x\right) + c$, where c is constant

Solution
For problems like

$\int \frac{\mathrm{dx}}{a + b x}$,

we start with assuming $a + b x = u$
then, differentiating this assumption $b \cdot \mathrm{dx} = \mathrm{du}$

$\mathrm{dx} = \frac{\mathrm{du}}{b}$

Now, substituting this in problem,

$\int \frac{\mathrm{du}}{b \cdot u} = \frac{1}{b} \cdot \ln u + c$, where c is constant

now, substituting u in the solution,

$\frac{1}{b} \cdot \ln \left(a + b x\right) + c$

Similarly following for the problem,
let $5 - 3 x = u$
then, $- 3 \cdot \mathrm{dx} = \mathrm{du}$
now substituting in the problem, we get

$\int - \frac{1}{3} \cdot \frac{\mathrm{du}}{u} = - \frac{1}{3} \cdot \ln \left(u\right) + c$

Finally, plugging in u, the answer will be $- \frac{1}{3} \cdot \ln \left(5 - 3 x\right) + c$