Question

$\int \frac{1}{(x^2-1)(x^2+2)}dx=$

• A. $\frac{1}{3}[\frac{1}{2}\log |\frac{x+1}{x-1}|-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}]+c$
• B. $\frac{1}{3}[\frac{1}{2}\log |\frac{x-1}{x+1}|+\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}]+c$
• C. $\frac{1}{3}[\frac{1}{2}\log |\frac{x-1}{x+1}|-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}]+c$
• D. $\frac{1}{3}[\frac{1}{3}\log |\frac{x-1}{x+1}|-\frac{1}{2\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}]+c$

Answer 1

The correct option is C $\frac{1}{3}[\frac{1}{2}\log |\frac{x-1}{x+1}|-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}]+c$

$\int \frac{1}{(x^2-1)(x^2+2)}dx{=}^1{/}_3\int \frac{(x^2+2)-(x^2-1)}{(x^2-1)(x^2+2)}dx$

$={}^1{/}_3[\int \frac{1}{(x^2-1)}dx-\int \frac{1}{(x^2+1)}dx]$

$={}^1{/}_3[{}^1{/}_2\int \frac{(x+1)-(x-1)}{(x^2-1)}dx-\int \frac{1}{(x^2+2)}dx]$

$={}^1{/}_3[{}^1{/}_2\left[log\right(x-1)-log(x+1\left)\right]-{}^1{/}_{\sqrt{2}}ta{n}^{-1}x{/}_{\sqrt{2}}]+C$

$={}^1{/}_3[{}^1{/}_{2}log\left|\frac{x-1}{x+1}\right|-1{/}_{\sqrt{2}}ta{n}^{-1}x{/}_{\sqrt{2}}]+C$