Question

If $I={\int }_2^3\frac{2x^5+x^4-2x^3+2x^2+1}{(x^2+1)(x^4-1)}dx$ then $I$equals

• A. $\frac{1}{2}\log 6+\frac{1}{10}$
• B. $\frac{1}{2}\log 6-\frac{1}{10}$
• C. $\frac{1}{2}\log 3-\frac{1}{10}$
• D. $\frac{1}{2}\log 2+\frac{1}{10}$

Answer 1

The correct option is B $\frac{1}{2}\log 6-\frac{1}{10}$

$I={\int }_2^3\frac{{2x}^5+x^4-{2x}^3+{2x}^2+1}{(x^2+1)(x^4-1)}dx$

As ${2x}^5+x^4-{2x}^3+{2x}^2+1$ $={2x}^3(x^2-1)+{(x^2+1)}^2$

$\therefore I={\int }_2^3\frac{{2x}^3(x^2-1)+{(x^2+1)}^2}{{(x^2+1)}^2\times (x^2-1)}dx$

$={\int }_2^3\frac{{2x}^3}{{(x^2+1)}^2}dx+{\int }_2^3\frac{dx}{x^2-1}$

$=I_1+{\left[\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|\right]}_2^3=I_1+\frac{1}{2}(\log \frac{1}{2}-\log \frac{1}{3})$

Where $I_1={\int }_2^3\frac{x^2}{{(x^2+1)}^2}\left(2x\right)dx$

Substitute $x^2+1=t\Rightarrow 2xdx=dt$

$\therefore I_1={\int }_5^{10}\frac{t-1}{t^2}dt={[\log \left|x\right|+\frac{1}{t}]}_5^{10}=\log 2-\frac{1}{10}$

$\therefore I=\frac{1}{2}\log 6-\frac{1}{10}$