How do you use Integration by Substitution to find #intdx/(5-3x)#?
1 Answer
Jul 25, 2014
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
then, differentiating this assumption
#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
then,
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