What is #f(x) = int (x-2)/((x+1)(x-4) ) dx# if #f(2)=5 #?

1 Answer
Nov 15, 2016

Use integration by partial fractions.

#A/(x + 1) + B/(x- 4) = (x - 2)/((x + 1)(x - 4))#

#A(x - 4) + B(x + 1) = x- 2#

#Ax - 4A + Bx + B = x - 2#

#(A + B)x + (B - 4A) = x - 2#

We can hence write the following system of equations.

#{(A + B = 1), (B - 4A = -2):}#

Solve:

#B = 1-A#

#1 - A - 4A = -2#

#-5A = -3#

#A = 3/5#

#A + B = 1#

#3/5 + B = 1#

#B = 2/5#

Hence, the partial fraction decomposition is #3/(5(x + 1)) + 2/(5(x - 4))#. We integrate using the rule #int(1/x)dx = ln|x| + C#.

#=>3/5ln|x + 1| + 2/5ln|x - 4| + C#

The function is #y= 3/5ln|x + 1| + 2/5ln|x - 4| + C#. We know an input/output of the function, so in this case we will solve for #C# to find the specific function.

We have that when #x =2#, #y = 5#.

#5 = 3/5ln|2 + 1| + 2/5ln|2 - 4| + C#

#5 = 3/5ln3 + 2/5ln2 + C#

#C = 5 - 3/5ln3 - 2/5ln2#

#C~=4.06#

#:.#The final function is #y = 3/5ln|x + 1| + 2/5ln|x - 4| + 4.06#, nearly.

Hopefully this helps!