# Calculate the definite integral of f2(x), if fa(x)=1/(|x-a|+3)? limits of integration are x=0 and x=3

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### 1 Answer

Since fa(x)=1/(|x-a|+3), then f2(x) = 1/(|x-2|+3).

We'll have to solve the definite integral of f2(x) = 1/(|x-2|+3)

Int dx/(|x-2|+3)

We'll discuss the absolute value:

|x-2| = x-2, for x>2

|x-2| = -x+2, for x<2

Since the limits of integration are x = 0 to x = 3, we'll split the interval in 2:

x = 0 to x = 2 and x = 2 to x = 3, since the function has different expressions over the ranges [0,2] and [2,3].

Int dx/(|x-2|+3) = Int dx/(5-x) + Int dx/(x+1)

Int dx/(5-x) = F(2) – F(0)

Int dx/(x+1) = F(3) – F(2)

But Int dx/(5-x) = -ln|x-5|

F(2) – F(0) = -ln|2-5| + ln|0-5| = -ln3 + ln 5 = ln (5/3)

Int dx/(x+1) = ln|x+1|

F(3) – F(2) = ln|3+1| - ln|2+1| = ln 4-ln3 = ln (4/3)

Int dx/(|x-2|+3) = ln (5/3) + ln (4/3) = ln 20/9

**The definite integral of the function f2(x) = 1/(|x-2|+3) is Int dx/(|x-2|+3) = ln 20/9.**