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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Subjective Medium Published on 17th 09, 2020
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