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Question

# $\int \frac{{x}^{3}+x+1}{{x}^{2}-1}dx$

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}I=\int \left(\frac{{x}^{3}+x+1}{{x}^{2}-1}\right)dx$ As Degree of Numerator is greater than Degree of Denominator we divide numerator by denominator ${x}^{2}-1\stackrel{x}{\overline{){x}^{3}+x+1}}\phantom{\rule{0ex}{0ex}}{x}^{3}-x\phantom{\rule{0ex}{0ex}}\overline{)-+}\phantom{\rule{0ex}{0ex}}2x+1\phantom{\rule{0ex}{0ex}}\therefore \frac{{x}^{3}+x+1}{{x}^{2}-1}=x+\frac{2x+1}{{x}^{2}-1}.....\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{3}+x+1}{{x}^{2}-1}=x+\frac{2x}{{x}^{2}-1}+\frac{1}{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}I=\int xdx+\int \frac{2x}{{x}^{2}-1}+\int \frac{dx}{{x}^{2}-{1}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}{x}^{2}-1=t\phantom{\rule{0ex}{0ex}}⇒2xdx=dt\phantom{\rule{0ex}{0ex}}\therefore I=\int xdx+\int \frac{dt}{t}+\int \frac{dx}{{x}^{2}-{1}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}+\mathrm{log}\left|t\right|+\frac{1}{2}\mathrm{log}\left|\frac{x-1}{x+1}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}+\mathrm{log}\left|{x}^{2}-1\right|+\frac{1}{2}\mathrm{log}\left|\frac{x-1}{x+1}\right|+C$

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