How do you evaluate the definite integral #int (x^2+2x)/(x+1)^2# from #[0, 1]#?

Answer 1

#1/2#

Before evaluating, we always have to find the integral. We look to simplify the integral as much as possible to see if anything can be cancelled.

#int_0^1 (x(x + 2))/(x + 1)^2#
We can't cancel anything, but a u-substitution would be effective. Let #u = x + 1#. Then #du = dx#. Also note that #x = u - 1# and #x + 2 = u + 1#. The integral therefore becomes:
#int_0^1 ((u - 1)(u + 1))/u^2 du#

Expand this:

#int_0^1 (u^2 - u + u - 1)/u^2 du#
#int_0^1 (u^2 - 1)/u^2 du#

Break into separate fractions.

#int_0^1 u^2/u^2 - 1/u^2 du#
#int_0^1 1 - 1/u^2 du#
#int_0^1 1 - u^-2 du#
You can now integrate this as #intx^ndx = x^(n + 1)/(n + 1) + C#, where #n != -1#.
#[u + 1/u]_0^1#
Reverse the substitution, since the initial variable wasn't #u#, it was #x#.
#[x + 1 + 1/(x + 1)]_0^1#
Evaluate using the second fundamental theorem of calculus, which states that #int_a^b F(x)dx = f(b) - f(a)#, where #f'(x) = F(x)# in all #[a, b]#.
#1 + 1 + 1/(1 + 1) - (0 + 1 + 1/(0 + 1))#
#2 + 1/2 - 2#
#1/2#

Hopefully this helps!

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Answer 2

To evaluate the definite integral (\int_{0}^{1} \frac{x^2 + 2x}{(x + 1)^2} , dx), you can first decompose the integrand using partial fraction decomposition. Then, integrate each term individually.

After integrating, evaluate the integral at the upper limit (1) and subtract the value obtained by evaluating the integral at the lower limit (0).

Let's proceed with the solution:

  1. Perform partial fraction decomposition: [ \frac{x^2 + 2x}{(x + 1)^2} = \frac{Ax + B}{x + 1} + \frac{C}{(x + 1)^2} ]

  2. Find the values of (A), (B), and (C) by equating numerators: [ x^2 + 2x = (Ax + B)(x + 1) + C \cdot 1 ]

  3. Solve for (A), (B), and (C).

  4. After obtaining the partial fraction decomposition, integrate each term: [ \int \frac{Ax + B}{x + 1} , dx + \int \frac{C}{(x + 1)^2} , dx ]

  5. Evaluate the integrals from 0 to 1.

  6. Subtract the value of the integral at the lower limit from the value at the upper limit to find the definite integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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