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

Question

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Answer 1 · 2014-07-25T20:54:28

The answer is

$-1/3*ln(5-3x)+c$ , where c is constant

Solution
For problems like

$int dx/(a+bx)$ ,

we start with assuming $a+bx=u$
then, differentiating this assumption $b*dx=du$

$dx=(du)/b$

Now, substituting this in problem,

$int(du)/(b*u) = 1/b*lnu +c$ , where c is constant

now, substituting u in the solution,

$1/b*ln(a+bx)+c$

Similarly following for the problem,
let $5-3x=u$
then, $-3*dx=du$
now substituting in the problem, we get

$int-1/3*(du)/u = -1/3*ln(u)+c$

Finally, plugging in u, the answer will be $-1/3*ln(5-3x)+c$

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