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Question Evaluate the integral ∫9dx(x+1)(x2+x)\int \frac{9 d x}{(x+1)(x^{2}+x)}. Consider the partial fraction decomposition 9(x+1)(x2+x)=9x−9x(x+1)2\frac{9}{(x+1)(x^{2}+x)}=\frac{9}{x}-\frac{9 x}{(x+1)^{2}}. Identify any errors in the work shown.

Studdy Solution

STEP 1

Assumptions
1. We are given the integral ∫9dx(x+1)(x2+x)\int \frac{9 dx}{(x+1)(x^2+x)} to evaluate.
2. We are also given a partial fraction decomposition attempt: 9(x+1)(x2+x)=Ax+Bx(x+1)2\frac{9}{(x+1)(x^2+x)} = \frac{A}{x} + \frac{Bx}{(x+1)^2} with A=9A=9 and B=−9B=-9.
3. We need to identify any errors in the partial fraction decomposition provided.

STEP 2

First, we need to factor the denominator correctly. The denominator is given as (x+1)(x2+x)(x+1)(x^2+x), which can be factored further since x2+xx^2+x has a common factor of xx.
x2+x=x(x+1)x^2+x = x(x+1)

STEP 3

Now we can rewrite the denominator using the factored form.
(x+1)(x2+x)=(x+1)x(x+1)=x(x+1)2 (x+1)(x^2+x) = (x+1)x(x+1) = x(x+1)^2

STEP 4

The correct form of the partial fraction decomposition should include all terms that are inversely proportional to each factor in the denominator. Since our denominator is x(x+1)2x(x+1)^2, we should have three terms in the decomposition: one for 1x\frac{1}{x}, one for 1x+1\frac{1}{x+1}, and one for 1(x+1)2\frac{1}{(x+1)^2}.

STEP 5

The correct form of the partial fraction decomposition should be:
9x(x+1)2=Ax+Bx+1+C(x+1)2\frac{9}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}

STEP 6

Identify the error in the given work: The partial fraction decomposition omitted a Cx+1\frac{C}{x+1} term.

STEP 7

Identify the error in the given work: The Bx(x+1)2\frac{Bx}{(x+1)^2} term should not have an xx in the numerator. The correct term should be Bx+1\frac{B}{x+1}.

STEP 8

We will not use the values of AA and BB provided in the shown work, as they were calculated based on an incorrect partial fraction decomposition.

STEP 9

To find the correct values of AA, BB, and CC, we will multiply both sides of the correct partial fraction decomposition by the common denominator x(x+1)2x(x+1)^2 and then equate the coefficients of corresponding powers of xx on both sides.

STEP 10

Multiply through by the common denominator:
9=A(x+1)2+Bx(x+1)+Cx 9 = A(x+1)^2 + Bx(x+1) + Cx

STEP 11

Expand the terms on the right side:
9=A(x2+2x+1)+Bx2+Bx+Cx 9 = A(x^2 + 2x + 1) + Bx^2 + Bx + Cx

STEP 12

Combine like terms:
9=(A+B)x2+(2A+B+C)x+A 9 = (A+B)x^2 + (2A+B+C)x + A

STEP 13

Since the left side of the equation is a constant (9), the coefficients of x2x^2 and xx on the right side must be zero, and the constant term must equal 9. This gives us a system of equations:
\begin{align*} A + B &= 0 \quad \text{(coefficient of } x^2\text{)} \\ 2A + B + C &= 0 \quad \text{(coefficient of } x\text{)} \\ A &= 9 \quad \text{(constant term)} \end{align*}

STEP 14

We already know that A=9A = 9 from the constant term equation. We can substitute this value into the other equations to find BB and CC.

STEP 15

Substitute A=9A = 9 into the first equation:
9+B=0 9 + B = 0

STEP 16

Solve for BB:
B=−9 B = -9

STEP 17

Substitute A=9A = 9 and B=−9B = -9 into the second equation:
2(9)−9+C=0 2(9) - 9 + C = 0

STEP 18

Solve for CC:
18−9+C=0 18 - 9 + C = 0
C=−9 C = -9

STEP 19

Now we have the correct values for AA, BB, and CC. The correct partial fraction decomposition is:
9x(x+1)2=9x−9x+1−9(x+1)2 \frac{9}{x(x+1)^2} = \frac{9}{x} - \frac{9}{x+1} - \frac{9}{(x+1)^2}

STEP 20

We can now integrate each term separately:
∫(9x−9x+1−9(x+1)2)dx \int \left( \frac{9}{x} - \frac{9}{x+1} - \frac{9}{(x+1)^2} \right) dx

STEP 21

Integrate the first term:
∫9xdx=9ln⁡∣x∣ \int \frac{9}{x} dx = 9 \ln|x|

STEP 22

Integrate the second term:
∫9x+1dx=9ln⁡∣x+1∣ \int \frac{9}{x+1} dx = 9 \ln|x+1|

STEP 23

Integrate the third term:
∫9(x+1)2dx=−9x+1 \int \frac{9}{(x+1)^2} dx = -\frac{9}{x+1}

STEP 24

Combine the results of the integrals:
9ln⁡∣x∣−9ln⁡∣x+1∣−9x+1+C 9 \ln|x| - 9 \ln|x+1| - \frac{9}{x+1} + C
where CC is the constant of integration.

STEP 25

The final answer is:
∫9dx(x+1)(x2+x)=9ln⁡∣x∣−9ln⁡∣x+1∣−9x+1+C \int \frac{9 dx}{(x+1)(x^2+x)} = 9 \ln|x| - 9 \ln|x+1| - \frac{9}{x+1} + C

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