using partial fractions, find `int (x^3+1)/(x(x-1)^3) dx`

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

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Using Partial fractions;

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

`x^3+1 = A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx`


x=0 we get `A=-1`

x=1 We get `D=2`

x=2 We get `9=A+2B+2C+2D rarr B+C=3`

x=-1 we get have `0=- 8A-4B+2C-D rarr 2B-C=3`


`2B-C=3 `

Solving the above two gives you;

`B = 2` and `C = 1`

`(x^3+1)/(x(x-1)^3) = -1/x + 2/(x-1) +1/(x-1)^2+2/(x-1)^3`

`int (x^3+1)/(x(x-1)^3) dx`

`= int(-1)/xdx + int 2/(x-1)dx +int1/(x-1)^2dx+int2/(x-1)^3dx`

`=-ln|x|+2ln|x-1|-1/(x-1)-1/(x-1)^2+C` Where C is a constant

So the answer is;

`int (x^3+1)/(x(x-1)^3) dx= -ln|x|+2ln|x-1|-1/(x-1)-1/(x-1)^2+C`


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