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Mathematics Question on Definite Integral

The value of ∫x2+1x2−1dx\int \frac{x^2+1}{x^2-1}dx is

A

Log(x+1x−1)+CLog \left (\frac{x+1}{x-1}\right)+C

B

Log(x−1x+1)+CLog \left(\frac{x-1}{x+1}\right)+C

C

Log(x2−1)+CLog (x^2 -1 )+C

D

x+Log(x−1x+1)+Cx+Log \left(\frac{x-1}{x+1}\right)+C

Answer

x+Log(x−1x+1)+Cx+Log \left(\frac{x-1}{x+1}\right)+C

Explanation

Solution

Let I=∫x2+1x2−1dxI = \int \frac{x^{2} +1}{x^{2} -1} dx
⇒I=∫x2+1−1+1x2−1dx\Rightarrow I = \int \frac{x^{2} + 1 -1 + 1}{x^{2} -1} dx
⇒I=∫x2−1x2−1dx+∫2x2−1dx\Rightarrow I = \int \frac{x^{2} -1}{ x^{2} -1} dx + \int \frac{2}{x^{2} -1} dx
⇒I=∫1dx+2∫1x2−1dx\Rightarrow I = \int 1 dx + 2 \int \frac{1}{x^{2} -1} dx
⇒I=x+212log⁡(x−1x+1)+c\Rightarrow I = x + 2 \frac{1}{2} \log \left(\frac{x-1}{x+1}\right) +c
⇒I=x+log⁡(x−1x+1)+c\Rightarrow I = x + \log \left(\frac{x-1}{x+1} \right) +c