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

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Answer 1 · 2018-05-07T13:52:03

#f(x) = 1+ (4 ln abs(x+5)+5 ln abs(x-4))/9 - (4 ln 7+5 ln 2)/9#

Decompose the rational function in partial fractions:

#(x+1)/((x+5)(x-4)) = A/(x+5)+B/(x-4)#

#(x+1)/((x+5)(x-4)) = (A(x-4)+B(x+5))/((x+5)(x-4))#

#x+1 = (A+B)x -4A+5B#

#{(A+B=1),(-4A+5B =1):}#

#{(A=4/9),(B=5/9):}#

Then:

#int (x+1)/((x+5)(x-4))dx = 4/9 intdx/(x+5)+5/9 int dx/(x-4)#

#int (x+1)/((x+5)(x-4))dx = 4/9 ln abs(x+5)+5/9 ln abs(x-4)+C#

Now, if #f(2) = 1# then:

#f(x) = 1+int_2^x (t+1)/((t+5)(t-4))dt#

#f(x) = 1+ (4 ln abs(x+5)+5 ln abs(x-4))/9 - (4 ln 7+5 ln 2)/9#