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
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