Question

How do you integrate `(2x+1 )/ ((x-2)(x^2+4))` using partial fractions?

Answer 1

`5/8ln|x-2|+3/8tan^-1(x/2)-5/16ln|x^2+4|+C`

Explanation:

We want to find `A`, `B`, and `C` such that `(2x+1)/((x-2)(x^2+4))=A/(x-2)+(Bx+C)/(x^2+4)`.

We multiply both sides by `(x-2)(x^2+4)` to get `2x+1=A(x^2+4)+(Bx+C)(x-2)`. Since we want to find `A`, `B`, and `C` such that all `x` satisfies the above equation, we can arbitrarily substitute values that simplify the problem, such as `x=2`.

Then, the equation would become `5=8A`. `A` then is `5/8`.

We can also substitute `x=0`: `1=4A-2C`. However, since `A=5/8`, `1=5/2-2C`. Solving for `C` gives `3/4`.

Knowing `A` and `C`, we can easily solve for `B` in this equation `2x+1=5/8(x^2+4)+(Bx+3/4)(x-2)` by setting `x=1` to get `B=-5/8`.

After substituting the values in and simplifying, we have successfully decomposed the fraction to `5/(8(x-2))+(-5x+6)/(8(x^2+4))`.

We can integrate the separate fractions one by one:

`\int\ 5/(8(x-2))\ dx`
`=5/8\int\ 1/u\ du` (substituting `u=x-2` and `du=dx`)
`=5/8ln|u|+C`
`=5/8ln|x-2|+C` (substituting `u=x-2` back)

`\int\ (-5x+6)/(8(x^2+4))\ dx`
`=-5/8\int\ (x-6/5)/(x^2+4)\ dx`
`=-5/8(\int\ x/(x^2+4)\ dx-\int\ (6/5)/(x^2+4)\ dx)`
`=-5/8(\int\ x/(2xu)\ du-\int\ (6/5)/(x^2+4)\ dx)` (substituting `u=x^2+4` and `du=2x\ dx`)
`=-5/8(1/2ln|u|+C-\int\ (6/5)/(x^2+4)\ dx)`
`=-5/8(1/2ln|x^2+4|+C-\int\ (6/5)/(x^2+4)\ dx)` (substituting `u=x^2+4` back)
`=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(x^2+4)\ dx)`
`=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(4(tan^2(\theta)+1))\ 2sec^2(\theta)\ d\theta)` (substituting `\2tan(\theta)=x` and `dx=2sec^2(\theta)\ d\theta`)
`=-5/8(1/2ln|x^2+4|+C-6/5\int\ 1/(4sec^2(\theta))\ 2sec^2(\theta)\ d\theta)`
`=-5/8(1/2ln|x^2+4|+C-3/5\theta)`
`=-5/8(1/2ln|x^2+4|+C-3/5tan^-1(x/2))` (since `x/2=tan(\theta)`)
`=3/8tan^-1(x/2)-5/16ln|x^2+4|+C`

Finally, we add the two integrals to get the final answer: `5/8ln|x-2|+3/8tan^-1(x/2)-5/16ln|x^2+4|+C`