The value of int frac{x^{2}+1}{x^{2}-1} d x i | vedclass

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(C) Let $I = \int \frac{x^{2}+1}{x^{2}-1} d x$. We can rewrite the numerator as $x^{2}-1+2$. So,$I = \int \frac{x^{2}-1+2}{x^{2}-1} d x$. $I = \int \l

Vedclass · Accepted Answer

$x+\log \left(\frac{x-1}{x+1}\right)+c$

Vedclass · Answer

(C) Let $I = \int \frac{x^{2}+1}{x^{2}-1} d x$. We can rewrite the numerator as $x^{2}-1+2$. So,$I = \int \frac{x^{2}-1+2}{x^{2}-1} d x$. $I = \int \left( \frac{x^{2}-1}{x^{2}-1} + \frac{2}{x^{2}-1} \right) d x$. $I = \int 1 d x + 2 \int \frac{1}{x^{2}-1} d x$. Using the standard integral formula $\int \frac{1}{x^{2}-a^{2}} d x = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + c$,where $a=1$: $I = x + 2 \cdot \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + c$. $I = x + \log \left| \frac{x-1}{x+1} \right| + c$.

The value of $\int \frac{x^{2}+1}{x^{2}-1} d x$ is

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