Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 52 Marks
Question
Evaluate: $\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x$
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Answer
Let $I =\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x$ where $\left(x^3-x^2-x+1\right)=x^2(x-1)-(x-1)=(x-1)\left(x^2-1\right)=(x-1)^2(x+1)$ Now let $\frac{3 x+5}{\left(x^3-x^2-x+1\right)}=\frac{3 x+5}{(x-1)^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$ $\Rightarrow(3 x +5)= A ( x -1)( x +1)+ B ( x +1)+ C ( x -1)^2$ Putting $x=1$ on both sides of (i), we get $B=4$ Putting $x=-1$ on both sides of (i), we get $C=\frac{1}{2}$ Comparing the coefficient of $x^2$ on both sides of ( $i$ ), we get $A+C=0 \Rightarrow A=-C=\frac{-1}{2}$ $\therefore \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}$ $\Rightarrow I=\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x=-\frac{1}{2} \int \frac{d x}{(x-1)}+4 \int \frac{d x}{(x-1)^2}+\frac{1}{2} \int \frac{d x}{(x+1)}$ $=-\frac{1}{2} \log |x-1|-\frac{4}{(x-1)}+\frac{1}{2} \log |x+1|+C$
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