What is the integral of the function (3x^2 +5x +6)/(x^3-16x). What are the steps to complete the functions integral

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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This can be done using partial fractions.

`(3x^2 +5x +6)/(x^3-16x) = (3x^2 +5x +6)/(x(x^2-16))`

`(3x^2 +5x +6)/(x^3-16x) = (3x^2 +5x +6)/(x(x^2-4^2))`

`(3x^2 +5x +6)/(x^3-16x) = (3x^2 +5x +6)/(x(x+4)(x-4))`

 

`(3x^2 +5x +6)/(x^3-16x) = A/x+B/(x-4)+C/(x+4)`

`(3x^2 +5x +6) = A(x-4)(x+4)+B(x)(x+4)+C(x-4)x`

`(3x^2 +5x +6) = Ax^2-16A+Bx^2+4Bx+Cx^2-4Cx`


Comparering components;

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

`x rarr 5 = 4(B-C) ----(2)`

`cons. rarr 6 = -16A ---(3)`

 

Solving the above 3 equations will give you;

`A = -3/8`

`B = 37/16`

`C = 17/16`

So we can say;

`(3x^2 +5x +6) = (-3/8)/x+(37/16)/(x-4)+(17/16)/(x+4)`

 

`int(3x^2 +5x +6)/(x^3-16x)dx`

`= int[(-3/8)/x+(37/16)/(x-4)+(17/16)/(x+4)]dx`

`= -3/8int1/xdx+37/16int1/(x-4)dx+17/16int1/(x+4)dx`

`= -3/8lnx+37/16ln(x-4)+17/16ln(x+4)+C ` where C is constant

 

So the answer is;

`int(3x^2 +5x +6)/(x^3-16x)dx = -3/8lnx+37/16ln(x-4)+17/16ln(x+4)+C`

 

 

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