Mathematics

# Find $\displaystyle \int \dfrac{x^{3}+x+1}{x^{2}-1}\ dx$

##### SOLUTION
$\int \frac{x^{3}x+1}{x^{2}-1}dx = \int (\frac{x^{3}}{x^{2}-1}+ \frac{x+1}{x^{2}-1})dx$
$=\int \frac{x^{3}dx}{x^{2}-1} + \int \frac{(x+1)dx}{(x+1)(x-1)}$
$= \int \frac{x^{3}dx}{x^{2}-1} + \int \frac{1 dx}{x-1}$
Now $\int \frac{x^{3}dx}{x^{2}-1}$
Put $x^{2}-1$
differentiating wrt.x
2x .dx = dt $\therefore xdx = \frac{dt}{2}$
$= \int \frac{x^{2}.xdx}{x^{2}-1} = \int \frac{t.dt}{2(t-1)}= \frac{1}{2}\int \frac{t}{t-1}$
$= \frac{1}{2}[\int \frac{t-1}{t-1}dt + \int \frac{1}{t-1}dt]$
$= \frac{1}{2}[\int dt + log(t-1)]$
$= \frac{t}{2} + \frac{1}{2}log(t-1)$
$= \frac{x^{2}}{2} + \frac{1}{2}log (x^{2}-1)+c$
$\therefore \int \frac{x^{3}+x+1}{x^{2}-1} = \int \frac{x^{3}dx}{x^{2}-1}+\int \frac{dx}{x-1}$
$= \frac{x^{2}}{2} + \frac{1}{2}log (x^{2}-1) + log(x-1) + c$
$= \frac{x^{2}}{2} + log\sqrt{x^{2}-1} + log (x-1) + c$
$= \frac{x^{2}}{2} + log (\sqrt{x^{2}-1}(x-1)) +c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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