Mathematics

Integrate:
$$\displaystyle\int{\dfrac{{{\left( x-1 \right)}^{2}}}{{{x}^{3}}-3{{x}^{2}}+3x+8}}dx$$


SOLUTION

Consider the given integral.


$$I=\displaystyle\int{\dfrac{{{\left( x-1 \right)}^{2}}}{{{x}^{3}}-3{{x}^{2}}+3x+8}}dx$$


 


Let $$t={{x}^{3}}-3{{x}^{2}}+3x+8$$


$$ \dfrac{dt}{dx}=3{{x}^{2}}-6x+3 $$


$$ \dfrac{dt}{3}={{x}^{2}}-2x+1 $$


$$ \dfrac{dt}{3}={{\left( x-1 \right)}^{2}} $$


 


Therefore,


$$ I=\dfrac{1}{3}\displaystyle\int{\dfrac{1}{t}}dt $$


$$ I=\dfrac{1}{3}\ln \left( t \right)+C $$


 


On putting the value of $$t$$, we get


$$I=\dfrac{1}{3}\ln \left( {{x}^{3}}-3{{x}^{2}}+3x+8 \right)+C$$


 


Hence, this is the answer.

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