Question

Show that $\int \frac{(x^2-1)dx}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}=\frac{1}{2}\log {\tan }^{-1}(x+\frac{1}{x})$.

Answer 1

Let $I=\int \frac{(x^2-1)dx}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}$
$=\int \frac{(1-\frac{1}{x^2})dx}{(x^2+3+\frac{1}{x^2}){\tan }^{-1}(x+\frac{1}{x})}$
$=\int \frac{(1-\frac{1}{x^2})dx}{({(x+\frac{1}{x})}^2+1){\tan }^{-1}(x+\frac{1}{x})}$
Put $x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{x^2})dx=dt$
Therefore
$I=\int \frac{dt}{(t^2+1){\tan }^{-1}t}$
Now put ${\tan }^{-1}t=u$

$\Rightarrow \frac{dt}{1+t^2}=du$

Hence
$I=\int \frac{du}{u}$

$I=\log u+C$

$=\log {\tan }^{-1}t+C$

$\Rightarrow I=\log {\tan }^{-1}(x+\frac{1}{x})+C$