Question

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

• A. $\frac{1}{5}[\log |x-2|-\frac{1}{2}\log (x^2+1)+2{\tan }^{-1}x]+c$
• B. $\frac{1}{5}[\log |x-2|-\frac{1}{2}\log (x^2+1)-2{\tan }^{-1}x]+c$
• C. $-\frac{1}{5}[\log |x-2|-\frac{1}{2}\log (x^2+1)-2{\tan }^{-1}x]+c$
• D. $-\frac{1}{2}[\log |x+2|+\frac{1}{2}\log |x^2-1|-2{\tan }^{-1}x]+c$

Answer 1

Answer: (B)

$\frac{1}{5}[\log |x-2|-\frac{1}{2}\log (x^2+1)-2{\tan }^{-1}x]+c$

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

$\frac{1}{(x-2)(x^2+1)}=\frac{A}{x-2}+\frac{Bx+c}{x^{2+1}}$

$1=A(x^2+1)+(Bx+c)(x-2)$

$A=\frac{1}{5};A+B=0-2B+C=0$

$B=\frac{-1}{5}C=\frac{-2}{5}$

$A=\frac{1}{5};B=\frac{-1}{5};C=\frac{-2}{5}$

$\int (\frac{\left[\frac{1}{5}\right]}{(x-2)}+\left[\frac{\left(\frac{-1}{5}\right)x-\left(\frac{-2}{5}\right)}{(x^2+1)}\right])dx$

$=\frac{1}{5}\left[log\right(x-2)-\frac{1}{2}log(x^2+1)-2ta{n}^{-1}x]+C$

So, $\int \frac{1}{(x-2)(x^2+1)}dx=\frac{1}{5}\left[log\right(x-2)-\frac{1}{2}log(x^2+1)-2ta{n}^{-1}x]+C$