How do you evaluate #\int \frac { x ^ { 3} + 2x ^ { 2} + 1} { ( x - 1) ( x - 2) } d x#?

Answer 1

#int (x^3+2x^2+1)/((x-1)(x-2))dx = x^2/2+5x-4lnabs(x-1)+17lnabs(x-2)+C#

A rational function must be broken down into its component parts using partial fractions in order to be integrated. However, in order to have a proper rational function with a numerator of lower degree than the denominator, polynomial division must be done first.

For the integrand, write:

#(x^3+2x^2+1)/((x-1)(x-2)) = (x^3+2x^2+1)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = (x^3-3x^2+2x+5x^2-2x+1)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = (x(x^2-3x+2)+5x^2-2x+1)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = x+(5x^2-2x+1)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = x+(5x^2-15x+10+13x-9)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = x+(5(x^2-3x+2)+13x-9)/(x^2-3x+2)#
#(x^3+2x^2+1)/((x-1)(x-2)) = x+5+(13x-9)/(x^2-3x+2)#

For the final fraction, use the partial fractions decomposition method:

#(13x-9)/(x^2-3x+2) = A/(x-1)+B/(x-2)#
#(13x-9)/(x^2-3x+2) = (A(x-2)+B(x-1))/((x-1)(x-2))#
#13x-9 = Ax-2A+Bx-B#
#{(A+B =13),(2A+B=9):}#
#{(A=-4),(B=17):}#

and lastly:

#(x^3+2x^2+1)/((x-1)(x-2)) = x+5-4/(x-1)+17/(x-2)#

and incorporating:

#int (x^3+2x^2+1)/((x-1)(x-2))dx = x^2/2+5x-4lnabs(x-1)+17lnabs(x-2)+C#
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Answer 2

To evaluate the integral (\int \frac{x^3 + 2x^2 + 1}{(x - 1)(x - 2)} dx), we first perform partial fraction decomposition.

We express the fraction (\frac{x^3 + 2x^2 + 1}{(x - 1)(x - 2)}) as the sum of two simpler fractions:

[\frac{x^3 + 2x^2 + 1}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}]

where (A) and (B) are constants to be determined.

Multiplying both sides by ((x - 1)(x - 2)), we get:

[x^3 + 2x^2 + 1 = A(x - 2) + B(x - 1)]

Expanding the right side:

[x^3 + 2x^2 + 1 = Ax - 2A + Bx - B]

Now, equating coefficients of like terms:

For (x^3): (A = 1)

For (x^2): (2 + B = 2 \implies B = 0)

For (x): (0 = -2A + B = -2(1) + 0 = -2 \implies A = 1)

Now that we have (A = 1) and (B = 0), we rewrite the integral as:

[\int \frac{1}{x - 1} dx + \int \frac{0}{x - 2} dx]

[\int \frac{1}{x - 1} dx + C]

Now, we can integrate each term separately:

[\int \frac{1}{x - 1} dx = \ln|x - 1| + C_1]

where (C_1) is the constant of integration.

Thus, the final result is:

[\int \frac{x^3 + 2x^2 + 1}{(x - 1)(x - 2)} dx = \ln|x - 1| + C]

where (C) is the constant of integration.

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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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