SOLUTION: Hello, Please help me to find the solution to the following: Compute the definite integral from 3 to 5 of f(x) dx, where {{{f(x)=(x)/(x-1)+(1)/(x+1)-x-(1)/(3)}}}. Many thanks.

Algebra ->  Functions -> SOLUTION: Hello, Please help me to find the solution to the following: Compute the definite integral from 3 to 5 of f(x) dx, where {{{f(x)=(x)/(x-1)+(1)/(x+1)-x-(1)/(3)}}}. Many thanks.      Log On


   

Question 1117638: Hello,
Please help me to find the solution to the following: Compute the definite integral from 3 to 5 of f(x) dx, where f%28x%29=%28x%29%2F%28x-1%29%2B%281%29%2F%28x%2B1%29-x-%281%29%2F%283%29.
Many thanks.
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
rewrite as a sum of integrals
:
integral(x/(x-1))dx + integral(1/(x+1))dx - integral(x)dx - (1/3)integral(1)dx
:
let u = x+1 and du = dx
:
integral(1/u)du + integral(x/(x-1))dx - integral(x)dx - (1/3)integral(1)dx
:
integral of 1/u is ln(u) where ln is he natural logarithm
:
ln(u) + integral(x/(x-1))dx - integral(x)dx - (1/3)integral(1)dx
:
replace x/(x-1) with its equivalent 1/(x-1) + 1
:
ln(u) + integral(1/(x-1) + 1)dx - integral(x)dx - (1/3)integral(1)dx
:
combine terms
:
ln(u) + integral(1/(x-1))dx +(2/3)integral(1)dx - integral(x)dx
:
substitute s = x-1 and ds = dx
:
ln(u) + integral(1/s)ds +(2/3)integral(1)dx - integral(x)dx
:
integral(1/s) is ln(s)
:
ln(u) + ln(s) +(2/3)integral(1)dx - integral(x)dx
:
integral of 1 is x and integral of x is x^2/2
:
ln(u) + ln(s) + (2/3)x - x^2/2
:
u = x+1 and s = x-1
:
ln(x+1) + ln(x-1) + (2/3)x - x^2/2
:
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ln(x^2-1) + (1/6)(4-3x)x
:
evaluate this expression for x = 5 and x = 3
:
ln(24) + (1/6)(4-15)5 = -5.9886
:
ln(8) + (1/6)(4-9)3 = -0.4206
:
-5.9886 -(-0.4206) = -5.568
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