What is the value of the definite integral of x^2/(x^3+1)^2 from x = 1 to x = 2?
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calendarEducator since 2008
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Let f(x) = x^2/(x^3+1)^2
We need to find the integral of f(x) from 1 to 2.
Let F(x) = Int f(x)
==> The definite integral is:
I = F(2) = F(1).
Let us determine F(x).
==> F(x) = Int = x^2/(x^3+1)^2 dx
Let us assume that x^3+1 = u ==> du 3x^2 dx
==> F(x) = Int (1/u^2) * du/3
= Int du/ 3u^2
= (1/3) Int u^-2 du = (1/3) u^-1/-1 = -1/3u
==> F(x) = -1/3u
Now we will substitute back u= x^3 +1
==> F(x) = -1/3(x^3+1)
==> F(2) = -1/3(8+1) = -1/27
==> F(1) = -1/3(1+1) = -1/6
==> I = -1/27 + 1/6 = -2/54 + 9/54 = 7/54
Then the definite integral is 7/54.
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
I'm sure you mean the value of the definite integral of x^2/(x^3+1)^2 from x = 1 to x = 2.
First we find the indefinite integral of x^2/(x^3+1)^2
Int [x^2/(x^3+1)^2 dx]
let u = x^3 + 1, du = 3x^2 dx or du/3 = x^2 dx
=> Int [ (1/3) * (1/ u^2) du]
=> 1/3 [ u^-2 du]
=> (1/3)(-1) u^-1 + C
substitute u with x^3 + 1
=> (-1/3)(1/ x^3 + 1) + C
For x = 2
(-1/3)(1/ x^3 + 1) + C = (-1/3)(1/9) + C
For x = 1
(-1/3)(1/ x^3 + 1) + C = (-1/3)(1/2) + C
The difference is (-1/3)(1/9) + C - (-1/3)(1/2) + C
=> (-1/3)(1/9 - (1/2)
=> (-1/3)( -7/18)
=> 7/54
The required value is 7/54
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