Evaluate the integral integrate of ((3)/(x^2-x-2))dx
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To integrate this, we need to separate the function into simpler functions using partial fractions, then integrate each function separately.
`int 3/{x^2-x-2}dx` factor denominator
`=int 3/{(x-2)(x+1)}dx` split into partial fractions
this means that `3/{(x-2)(x+1)}=A/{x-2}+B/{x+1}` . Multiplying out and comparing on the constant terms and the linear terms, we see that:
`3=A(x+1)+B(x-2)` which gives the two equations:
`3=A-2B` equation (1)
`0=A+B` equation (2)
Subtract (1) from (2)
`3B=3` so B=1 and A=-1
Now the integral becomes
`-int 1/{x-2} dx + int1/{x+1}dx` integrate each term
`=-ln(x-2)+ln(x+1)+C` where C is the constant of integration
`=ln({x+1}/{x-2})+C`
The integral evaluates to `ln({x+1}/{x-2})+C` .
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