1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Evaluate the following integrals: $\int \frac{{x}^{3}+x+1}{{x}^{2}-1}\mathrm{d}x$$\int \frac{{x}^{3}+x+1}{{x}^{2}-1}\mathrm{d}x$

Open in App
Solution

## $\mathrm{Let}I=\int \frac{{x}^{3}+x+1}{{x}^{2}-1}\mathrm{d}x\phantom{\rule{0ex}{0ex}}\mathrm{Here}\mathrm{the}\mathrm{integrand}\frac{{x}^{3}+x+1}{{x}^{2}-1}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{proper}\mathrm{rational}\mathrm{function},\mathrm{so}\mathrm{we}\mathrm{divide}{x}^{3}+x+1\mathrm{by}{x}^{2}-1\mathrm{and}\mathrm{find}\mathrm{that}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{{x}^{3}+x+1}{{x}^{2}-1}=x+\frac{2x+1}{{x}^{2}-1}=x+\frac{2x+1}{\left(x+1\right)\left(x-1\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\frac{2x+1}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}\phantom{\rule{0ex}{0ex}}⇒2x+1=A\left(x-1\right)+B\left(x+1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}2=A+B\mathrm{and}1=-A+B\phantom{\rule{0ex}{0ex}}\mathrm{or}A=\frac{1}{2}\mathrm{and}B=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore I=\int \left(x+\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2}}{x-1}\right)\mathrm{d}x\phantom{\rule{0ex}{0ex}}=\int x\mathrm{d}x+\frac{1}{2}\int \frac{1}{x+1}\mathrm{d}x+\frac{3}{2}\int \frac{1}{x-1}\mathrm{d}x\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}+\frac{1}{2}\mathrm{log}\left|x+1\right|+\frac{3}{2}\mathrm{log}\left|x-1\right|+c\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}+\frac{1}{2}\mathrm{log}\left|x+1\right|+\frac{3}{2}\mathrm{log}\left|x-1\right|+c\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\int \frac{{x}^{3}+x+1}{{x}^{2}-1}\mathrm{d}x=\frac{{x}^{2}}{2}+\frac{1}{2}\mathrm{log}\left|x+1\right|+\frac{3}{2}\mathrm{log}\left|x-1\right|+c$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Special Integrals - 2
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program