How do you integrate #int (2x+4)/((x+2)(x-6)(x^2+3)) dx# using partial fractions?

Answer 1

#int(2x+4)/((x+2)(x-6)(x²+3))dx=1/39ln(x-2/13)-1/78ln(x²+3)-2/(13sqrt6)arctan(x/sqrt6)+C#,#C in RR#

#int(2x+4)/((x+2)(x-6)(x²+3))dx# #=int(2(cancel(x+2)))/((cancel(x+2))(x-6)(x²+3))dx# #=2int1/((x-6)(x²+3))dx# Now let #1/((x-6)(x²+3))=A/(x-6)+(Bx+C)/(x²+3)# #1/(cancel((x-6)(x²+3)))=(Ax²+3A+Bx²-6Bx+Cx-6C)/(cancel((x-6)(x²+3)))# #(A+B)x²+(C-6B)x+3A-6C=1# By identification : #A+B=0#[1] #C-6B=0#[2] #3A-6C=1#[3]
#A+B=0#[1] #6A+C=0#[2] #6A-12C=2#[3]
#A+B=0#[1] #6A+C=0#[2] #-13C=2#[3]
#A+B=0#[1] #6A-2/13=0#[2] #C=-2/13#[3]
#B=-1/39#[1] #A=1/39#[2] #C=-2/13#[3]

So:

#int(2x+4)/((x+2)(x-6)(x²+3))dx=int(1/(39(x-6))+((-1/39)x-2/13)/(x²+3))dx#
#=int(1/39(1/(x-2/13))dx-int(1/39x+2/13)/(x²+3)dx#
#=1/39int1/(x-2/13)dx-1/78int(2x/(x²+3))dx-1/39int(6/(x²+3))dx#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/13int(1/(x²+3))dx#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/13*1/sqrt6arctan(x/sqrt3)#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/(13sqrt6)arctan(x/sqrt6)+C#,#C in RR#

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Answer 2

To integrate ( \frac{2x+4}{(x+2)(x-6)(x^2+3)} ) using partial fractions, follow these steps:

  1. Factor the denominator completely: ( (x+2)(x-6)(x^2+3) ).
  2. Write the fraction in the form of partial fractions: ( \frac{2x+4}{(x+2)(x-6)(x^2+3)} = \frac{A}{x+2} + \frac{B}{x-6} + \frac{Cx+D}{x^2+3} ).
  3. Clear the denominators by multiplying both sides by the common denominator ( (x+2)(x-6)(x^2+3) ).
  4. After clearing denominators, equate the numerators: ( 2x+4 = A(x-6)(x^2+3) + B(x+2)(x^2+3) + (Cx+D)(x+2)(x-6) ).
  5. Substitute suitable values of ( x ) to solve for ( A ), ( B ), ( C ), and ( D ).
  6. Once you find the values of ( A ), ( B ), ( C ), and ( D ), rewrite the original integral using the partial fraction decomposition.
  7. Integrate each partial fraction separately.
  8. Finally, combine the integrals to obtain the result.
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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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