# Evaluate the integral. int(x^2-1)/(x^2)dx

tabita57i 2020-11-09 Answered
Evaluate the integral.
$\int \frac{{x}^{2}-1}{{x}^{2}}dx$
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## Expert Answer

2k1enyvp
Answered 2020-11-10 Author has 94 answers
Given That
$\int \frac{{x}^{2}-1}{{x}^{2}}dx$
$=\int \left(1-\frac{1}{{x}^{2}}\right)dx$
$=\int dx-\int \frac{1}{{x}^{2}}dx$
$=x-\left[\frac{{x}^{-2+1}}{-2+1}\right]+c$
$=x+\frac{1}{x}+c$
$\therefore \int \frac{{x}^{2}-1}{{x}^{2}}dx=x+\frac{1}{x}+c$
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