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

Answer 1

#=ln((x^2+1)/x)+C#

#(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)# Developing #(x^2-1)/(x(x^2+1))=(A(x^2+1)+ x(Bx+C))/(x(x^2+1)# So we can compare the coefficients #x^2-1=A(x^2+1)+ x(Bx+C)# For #x^2# , #1=A+B# and #-1=A# and #0=C# Hence #A=-1#and #B=2# #int((x^2-1)dx)/(x(x^2+1))=intAdx/x+int((Bx+C)dx)/(x^2+1)# #=int(-1dx)/x+int((2x)dx)/(x^2+1# #=-lnx +ln(x^2+1)+C# #=ln((x^2+1)/x)+C#
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Answer 2

To integrate the function ( \frac{{x^2 - 1}}{{x(x^2 + 1)}} ) using partial fractions, first factor the denominator ( x(x^2 + 1) ) as ( x \cdot (x + i)(x - i) ) where ( i ) is the imaginary unit. Then express the fraction as the sum of two simpler fractions:

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

Multiply both sides by the denominator ( x(x^2 + 1) ) to clear the fractions:

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

Expand and equate coefficients:

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

Now, equate coefficients of like terms:

For ( x^2 ): ( A + B = 1 ) For ( x ): ( C = 0 ) For constant term: ( A = -1 )

Solve the system of equations to find the values of ( A ) and ( B ):

[ A = -1 ] [ A + B = 1 \implies -1 + B = 1 \implies B = 2 ]

Now, we have:

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

Integrate each term separately:

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

[ \int{\frac{{2x}}{{x^2 + 1}} , dx} = \ln{|x^2 + 1|} + C_2 ]

So, the integral of ( \frac{{x^2 - 1}}{{x(x^2 + 1)}} ) using partial fractions is:

[ -\ln{|x|} + \ln{|x^2 + 1|} + C ]

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Answer 3

To integrate (\int \frac{x^2 - 1}{x(x^2 + 1)} , dx) using partial fractions, follow these steps:

  1. Factor the denominator, (x(x^2 + 1)), if possible.

  2. Express the rational function as a sum of simpler fractions using partial fraction decomposition.

  3. Solve for the coefficients of the partial fractions.

  4. Integrate each partial fraction.

  5. Combine the results to obtain the final integral.

Here's how to apply these steps:

  1. Factor the denominator: (x(x^2 + 1)) cannot be factored further.

  2. Express the rational function as partial fractions: [ \frac{x^2 - 1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} ]

  3. Solve for the coefficients (A), (B), and (C). Multiply through by the denominator to clear the fractions: [ x^2 - 1 = A(x^2 + 1) + (Bx + C)x ] Expanding and comparing coefficients, you get the following equations: [A = 1] [B = -1] [C = 0]

  4. Integrate each partial fraction: [ \int \frac{1}{x} , dx + \int \frac{-x}{x^2 + 1} , dx ] Integrating each term yields: [ \ln|x| - \frac{1}{2} \ln|x^2 + 1| + C ]

  5. Combine the results: [ \int \frac{x^2 - 1}{x(x^2 + 1)} , dx = \ln|x| - \frac{1}{2} \ln|x^2 + 1| + C ]

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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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