How do you integrate #(x-1)/ (x^3 +x^2)# using partial fractions?

Answer 1
We find that, the #Dr.=x^3+x=x(x^2+1)#
This, the poly. of the #Dr.# has a linear factor #x#, and a non-reducible quadr. factor #x^2+1#. Therefore, we will suppose that
#(x-1)/(x^3+x)=(x-1)/(x(x^2+1))=A/x+(Bx+c)/(x^2+1), A,B,C in RR...(star)#.

However, after simplifying,

#The R.H.S.=(A(x^2+1)+(Bx+C)x)/(x(x^2+1))#.
Thus, #(x-1)/(x^3+x)=(A(x^2+1)+(Bx+C)x)/(x(x^2+1))... ......(1)#.
As the #Drs.# of #(1)# are same, so must be the #Nrs.#. Hence,
#Ax^2+A+Bx^2+Cx=(A+B)x^2+Cx+A=x-1......(2)#.

When we compare the resp. co-effs. of the two sides, we obtain,

#A=-1, C=1, and, A+B=0rArrB=-A=1#.
Therefore, by #(star)#,
#int(x-1)/(x^3+x)dx=int[-1/x+(x+1)/(x^2+1)]dx#
#=-int1/xdx+int(x+1)/(x^2+1)dx=-ln|x|+int{x/(x^2+1)+1/(x^2+1)}dx#
#=-ln|x|+1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx#
#=-ln|x|+1/2int(d/dx(x^2+1))/(x^2+1)dx+arctanx#
#=-ln|x|+1/2ln(x^2+1)+arctanx+C#, OR,
#=ln(sqrt(x^2+1)/|x|)+arctanx+C#.

This well-known Rule has been applied in the last step:

The Rule# := int(f'(x))/f(x)dx=ln|f(x)+K#.

I hope this helps! Have fun with math!

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Answer 2

To integrate the given expression using partial fractions, first, factor the denominator:

[ x^3 + x^2 = x^2(x + 1) ]

Now, we express the integrand as a sum of partial fractions:

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

Multiplying both sides by the denominator ( x^3 + x^2 ) to clear the fractions:

[ x - 1 = A(x + 1) + Bx(x + 1) + Cx^2 ]

Now, equating coefficients:

For ( x ): [ 1 = A + B ]

For ( x^2 ): [ 0 = C + B ]

For the constant term: [ -1 = A ]

Solving these equations, we find: [ A = -1 ] [ B = 1 ] [ C = -1 ]

So, the partial fraction decomposition is: [ \frac{x - 1}{x^3 + x^2} = \frac{-1}{x} + \frac{1}{x^2} - \frac{1}{x + 1} ]

Now, we can integrate each term separately. The integrals are: [ \int \frac{-1}{x} dx = -\ln|x| ] [ \int \frac{1}{x^2} dx = -\frac{1}{x} ] [ \int \frac{-1}{x + 1} dx = -\ln|x + 1| ]

Therefore, integrating the original expression using partial fractions yields: [ -\ln|x| - \frac{1}{x} - \ln|x + 1| + C ] where ( C ) is the constant of integration.

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Answer 3

To integrate (x - 1) / (x^3 + x^2) using partial fractions, follow these steps:

  1. Factor the denominator if possible. In this case, the denominator can be factored as x^2(x + 1).

  2. Decompose the rational function into partial fractions. Assume the decomposition takes the form:

    (x - 1) / (x^3 + x^2) = A / x + B / x^2 + C / (x + 1)

  3. Clear the fractions by multiplying both sides by the denominator:

    (x - 1) = A(x^2)(x + 1) + B(x + 1) + C(x)(x^2)

  4. Expand and equate coefficients of like terms.

  5. Once you find the values of A, B, and C, rewrite the original integral using the partial fraction decomposition.

  6. Integrate each term separately.

  7. If needed, simplify the resulting expression.

By following these steps, you can integrate the given rational function using partial fractions.

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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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