Question

If $I=\int {\cot }^{-1}\left(\frac{a^2-ax+x^2}{a^2}\right)dx$, then $I$ equals

• A. $x{\tan }^{-1}\left(\frac{x}{a}\right)-(x-a){\tan }^{-1}\left(\frac{x-a}{a}\right)+C$
• B. $\frac{a}{2}\log (2a^2-2ax+x^2)-\frac{a}{2}\log (x^2+a^2)+C$
• C. $x{\tan }^{-1}\left(\frac{x}{a}\right)+(x-a){\tan }^{-1}\left(\frac{x-a}{a}\right)+\frac{a}{2}\log (2a^2-2ax+x^2)+C$
• D. none of these

Answer 1

The correct option is D none of these

As $\frac{a^2-ax+x^2}{a^2}=1-\frac{x}{a}\left(\frac{a-x}{a}\right)$ and ${\cot }^{-1}x={\tan }^{-1}\left(\frac{1}{x}\right)$
${\cot }^{-1}\left(\frac{a^2-ax+x^2}{a^2}\right)={\tan }^{-1}\left[\frac{\frac{x}{a}+\left(\frac{a-x}{a}\right)}{1+\frac{x}{a}\left(\frac{a-x}{a}\right)}\right]={\tan }^{-1}\left(\frac{x}{a}\right)+{\tan }^{-1}\left(\frac{a-x}{a}\right)$
$\Rightarrow I=I_1+I_2$
Where
$I_1=\int {\tan }^{-1}\left(\frac{x}{a}\right)dx=x{\tan }^{-1}\left(\frac{x}{a}\right)-\int \frac{xa}{x^2+a^2}dx=x{\tan }^{-1}\left(\frac{x}{a}\right)-\frac{a}{2}\log (x^2+a^2)$
and
$I_2=\int {\tan }^{-1}\left(\frac{a-x}{a}\right)dx$
Put $\frac{a-x}{a}=t$
$I_2=-a\int {\tan }^{-1}tdt=-a[t{\tan }^{-1}t-\frac{1}{2}\log (t^2+1)]=(x-a){\tan }^{-1}\left(\frac{a-x}{a}\right)+\frac{a}{2}\log (2a^2-2ax+x^2)$
Hence no option