integral sign 1 over 0 (x^2 + 2x) / (x+1)^2 dx
Do you change the top to the complete the square form: (x-a)^2 + b or is there any other way that I would need to know about?
Do you change the top to the complete the square form: (x-a)^2 + b or is there any other way that I would need to know about?
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Divide
..................1
.................____________________
x² + 2x + 1|x² + 2x
..................x² + 2x + 1
.................----------------
..............................–1
So the quotient is 1 – 1/(x + 1)²
The integral of 1 is x, and the integral of –(x + 1)^(-2) is 1/(x + 1)
x + 1/(x + 1) from 0 to 1 is 1 + 1/2 – (0 + 1) = 1/2
..................1
.................____________________
x² + 2x + 1|x² + 2x
..................x² + 2x + 1
.................----------------
..............................–1
So the quotient is 1 – 1/(x + 1)²
The integral of 1 is x, and the integral of –(x + 1)^(-2) is 1/(x + 1)
x + 1/(x + 1) from 0 to 1 is 1 + 1/2 – (0 + 1) = 1/2
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(x^2 + 2x) / (x + 1)^2 =>
A + B/(x + 1) + C/(x + 1)^2
A * (x + 1)^2 + B * (x + 1) + C = x^2 + 2x + 0
A * (x^2 + 2x + 1) + Bx + B + C = x^2 + 2x + 0
Ax^2 + 2Ax + Bx + A + B + C = x^2 + 2x + 0
Ax^2 = x^2
A = 1
2Ax + Bx = 2x
2A + B = 2
2 * 1 + B = 2
2 + B = 2
B = 0
A + B + C = 0
1 + 0 + C = 0
C = -1
Now we have:
1 - 1/(x + 1)^2
Can we integrate that?
1 * dx - dx / (x + 1)^2
1 * dx is easy enough to integrate. x
Integrating -dx / (x + 1)^2 is a little trickier, but not much trickier
u = x + 1
du = dx
-du / u^2
Integrate with the power rule:
1/u + C =>
1/(x + 1) + C
Now we have:
x + 1/(x + 1) + C
From 0 to 1
1 + 1/(1 + 1) - 0 - 1/(0 + 1) =>
1 + 1/2 - 0 - 1/1 =>
1 + 1/2 - 1 =>
1/2
A + B/(x + 1) + C/(x + 1)^2
A * (x + 1)^2 + B * (x + 1) + C = x^2 + 2x + 0
A * (x^2 + 2x + 1) + Bx + B + C = x^2 + 2x + 0
Ax^2 + 2Ax + Bx + A + B + C = x^2 + 2x + 0
Ax^2 = x^2
A = 1
2Ax + Bx = 2x
2A + B = 2
2 * 1 + B = 2
2 + B = 2
B = 0
A + B + C = 0
1 + 0 + C = 0
C = -1
Now we have:
1 - 1/(x + 1)^2
Can we integrate that?
1 * dx - dx / (x + 1)^2
1 * dx is easy enough to integrate. x
Integrating -dx / (x + 1)^2 is a little trickier, but not much trickier
u = x + 1
du = dx
-du / u^2
Integrate with the power rule:
1/u + C =>
1/(x + 1) + C
Now we have:
x + 1/(x + 1) + C
From 0 to 1
1 + 1/(1 + 1) - 0 - 1/(0 + 1) =>
1 + 1/2 - 0 - 1/1 =>
1 + 1/2 - 1 =>
1/2
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The given integral may be written as:
∫ (x^2+2x) dx /(x^2+2x+1)
= ∫ [(x^2+2x+1)-1] dx /(x^2+2x+1)
∫ (x^2+2x) dx /(x^2+2x+1) = ∫ dx - ∫ dx /(x+1)^2 ---------(1)
∫ dx = x
∫ dx /(x+1)^2
Let u=(x+1)
du=dx
∫ dx /(x+1)^2 = ∫ du /u^2 = ∫ u^(-2) du = u^(-2+1)/(-2+1) = -1/u = -1/(x+1) ---(2)
Plug (2) into (1)
∫ (x^2+2x) dx /(x^2+2x+1) = x - (-1/(x+1))
= x+ 1/(x+1)
Substitute x= 0
x+1/(x+1) = 1
Substitute x= 1
x+1/(x+1) = 1+1/2 = 3/2
Subtract:
3/2 -1 = 1/2
∫ (x^2+2x) dx /(x^2+2x+1)
= ∫ [(x^2+2x+1)-1] dx /(x^2+2x+1)
∫ (x^2+2x) dx /(x^2+2x+1) = ∫ dx - ∫ dx /(x+1)^2 ---------(1)
∫ dx = x
∫ dx /(x+1)^2
Let u=(x+1)
du=dx
∫ dx /(x+1)^2 = ∫ du /u^2 = ∫ u^(-2) du = u^(-2+1)/(-2+1) = -1/u = -1/(x+1) ---(2)
Plug (2) into (1)
∫ (x^2+2x) dx /(x^2+2x+1) = x - (-1/(x+1))
= x+ 1/(x+1)
Substitute x= 0
x+1/(x+1) = 1
Substitute x= 1
x+1/(x+1) = 1+1/2 = 3/2
Subtract:
3/2 -1 = 1/2
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The solutions posted so far are OK but
complicated. Here's a simpler way:
Add and subtract 1 in the numerator.
Then you have
∫[0..1] (1 -1/(x+1)^2) dx
= x + 1/(x+1)[0..1] = 1 + 1/2 - 1 = 1/2.
complicated. Here's a simpler way:
Add and subtract 1 in the numerator.
Then you have
∫[0..1] (1 -1/(x+1)^2) dx
= x + 1/(x+1)[0..1] = 1 + 1/2 - 1 = 1/2.
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∫(x² + 2x)/(x + 1)² dx from 0 to 1
∫(x² + x + x)/(x + 1)² dx
∫x/(x + 1) dx + ∫x/(x + 1)² dx
∫(x + 1 - 1)/(x + 1) dx + ∫(x + 1 - 1)/(x + 1)² dx
∫dx - ∫dx/(x + 1)²
= x + 1/(x + 1) eval. from 0 to 1
= 1 + 1/2 - 1 = 1/2
∫(x² + x + x)/(x + 1)² dx
∫x/(x + 1) dx + ∫x/(x + 1)² dx
∫(x + 1 - 1)/(x + 1) dx + ∫(x + 1 - 1)/(x + 1)² dx
∫dx - ∫dx/(x + 1)²
= x + 1/(x + 1) eval. from 0 to 1
= 1 + 1/2 - 1 = 1/2