Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 23 Marks
Question
Evaluate: $\int \frac{x^3+x^2+2 x+1}{x^2-x+1} d x$
✓
Answer
Let $I =\int \frac{x^3+x^2+2 x+1}{x^2-x+1} d x$
Here we have,
$\frac{x^3+x^2+2 x+1}{x^2-x+1}= x +2+\frac{3 x-1}{x^2-x+1} \ldots (i)$
$\text { Let } 3 x -1=A \frac{d}{d x}( x - x +1)+ B$
$\Rightarrow 3 x -1= A (2 x +1)+ B$
$\Rightarrow 3 x -1=(2 A) x + B - A $
Equating Coefficients of like terms, we get
$2 A=3 \Rightarrow A=\frac{3}{2}$
$\text { Also, } B - A =-1$
$\Rightarrow B-\frac{3}{2}=-1$
$\Rightarrow B=\frac{1}{2}$
So, $\int\left(\frac{x^3+x^2+2 x+1}{x^2-x+1}\right) d x=\int( x +2) dx +\int\left(\frac{\frac{3}{2}(2 x-1)+\frac{1}{2}}{x^2-x+1}\right)$
$=\int( x +2) dx +\frac{3}{2} \int\left(\frac{2 x-1}{x^2-x+1}\right) d x+\frac{1}{2} \int \frac{d x}{x^2-x+1}$
$=\int( x +2) dx +\frac{3}{2} \int \frac{(2 x-1) d x}{x^2-x+1}+\frac{1}{2} \int \frac{d x}{x^2-x+\frac{1}{4}-\frac{1}{4}+1}$
$=\int( x +2) dx +\frac{3}{2} \int \frac{(2 x-1) d x}{x^2-x+1}+\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{x^2}{2}+2 x +\frac{3}{2} \log \left| x ^2- x +1\right|+\frac{1}{2} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$
$=\frac{x^2}{2}+2 x +\frac{3}{2} \log \left| x ^2- x +1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+C$
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