Question

If $\int \frac{(x-1{)}^2}{x^4+x^2+1}dx=\frac{1}{\sqrt{a}}{\tan }^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right)-\frac{b}{\sqrt{a}}{\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+C,$ then which of the following is/are CORRECT?
(where $a,b$ are fixed constants and $C$ is integration constant)

• A. $a^2+b^2=13$
• B. $a^2-b^2=13$
• C. $ab=6$
• D. $ab=12$

Answer 1

Answer: (A), (C)

$ab=6$

$I=\int \frac{(x-1{)}^2}{x^4+x^2+1}dxI=\int \frac{x^2+1}{x^4+x^2+1}dx-\int \frac{2x}{x^4+x^2+1}dx$
Dividing $N^r$ and $D^r$ by $x^2$
$=\underset{I_1}{\int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx}-\underset{I_2}{\int \frac{2x}{(x^2{)}^2+x^2+1}}\therefore I_1=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+3}dx$
Let $x-\frac{1}{x}=t$
$\Rightarrow (1+\frac{1}{x^2})dx=dt\Rightarrow I_1=\int \frac{dt}{t^2+(\sqrt{3}{)}^2}\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}\Rightarrow I_1=\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{t}{\sqrt{3}}+C=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C\cdots \left(i\right)$
and $I_2=\int \frac{2x}{(x^2{)}^2+x^2+1}dx$
Put $x^2=p$
$\Rightarrow 2xdx=dp{I}_2=\int \frac{dp}{p^2+p+1}=\int \frac{dp}{{(p+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}{I}_2=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2(x^2+\frac{1}{2})}{\sqrt{3}}\right)\cdots \left(ii\right)$
$\therefore I=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right)-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+C\therefore a=3,b=2$
Therefore,
$a^2+b^2=9+4=13$
$a^2-b^2=9-4=5$
$ab=6$