How do you integrate #int x^3/((x^4-16)(x-3))dx# using partial fractions?

Answer 1

#int (x^3)/((x^4-16)(x-3))=(-3)/52ln(x^2+4)+1/13tan^-1 (x/2)-1/20ln(x+2)-1/4ln(x-2)+27/65ln(x-3)+C_0#

Start from the given integral and set up the variables using A, B, C ,D, E

#int (x^3)/((x^4-16)(x-3))dx=#
#int((Ax+B)/(x^2+4)+C/(x+2)+D/(x-2)+E/(x-3))dx#

and out of the numerators from both sides we have

#x^3=Ax^4-4Ax^2-3Ax^3+12Ax+Bx^3-4Bx-3Bx^2+12B+Cx^4-5Cx^3+10Cx^2-20Cx+24C+Dx^4-Dx^3-2Dx^2-4Dx-24D+Ex^4-16E#

After setting up all the equations . The following are the equations

#C+D+E=0" "#equation 1 #B-5C-D=1" "#equation 2 #-3B+10C-2D=0" "#equation 3 #-4B-20C-4D=0" "#equation 4 #12B+24C-24D-16E=0" "#equation 5
Solving for A, B, C, D, E by any method #A=(-3)/26# #B=2/13# #C=(-1)/20# #D=(-1)/4# #E=27/65#

We have to integrate now

#int (x^3)/((x^4-16)(x-3))dx=#
#int(((-3)/26x+2/13)/(x^2+4)+(-1)/20/(x+2)+(-1)/4/(x-2)+27/65/(x-3))dx#
#int (x^3)/((x^4-16)(x-3))=(-3)/52ln(x^2+4)+1/13tan^-1 (x/2)-1/20ln(x+2)-1/4ln(x-2)+27/65ln(x-3)+C_0#
where #C_0# be the constant of integration

God bless....I hope the explanation is useful.

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Answer 2

To integrate the expression ( \int \frac{x^3}{(x^4-16)(x-3)} , dx ) using partial fractions, you first factor the denominator completely. After factoring, you decompose the fraction into partial fractions. You then solve for the unknown constants using algebraic manipulation. Once you have the partial fraction decomposition, you can integrate each term separately.

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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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