Question

The value of ${\int }_0^3\frac{{\tan }^{-1}(x-[x])}{1+(x-[x]{)}^2}dx$ is equal to [Note:$\left[y\right]$ denotes greatest integer function of $y$.]

• A. $0$
• B. $3$
• C. $\frac{3{\pi }^2}{32}$
• D. $\frac{3{\pi }^2}{16}$

Answer 1

The correct option is C $\frac{3{\pi }^2}{32}$

${\int }_0^1\frac{{\tan }^{-1}(x-[x\left]\right)}{1+(x-[x]{)}^2}$

${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx+{\int }_1^2\frac{{\tan }^{-1}(x-1)}{1+(x-1{)}^2}dx+{\int }_2^3\frac{{\tan }^{-1}(x-2)}{1+(x-2{)}^2}dx$

$u={\tan }^{-1}x;du=\frac{1}{1+x^2}dx$

$v={\tan }^{-1}(x-1);mdv=\frac{1}{1+(x-1{)}^2}dx$

$\int udu+\int vdv+\int wdw$

$=\frac{u^2}{2}+\frac{v^2}{2}+\frac{w^2}{2}$

${\frac{({\tan }^{-1}x{)}^2}{2}]}_0^1+{\frac{\left({\tan }^{-1}\right(x-1{)}^2)}{2}]}_1^2+{\frac{\left({\tan }^{-1}\right(x-2){)}^2}{2}]}_2^3$

$\frac{1}{2}[{\left(\frac{n}{4}\right)}^2+{\left(\frac{n}{4}\right)}^2+{\left(\frac{n}{4}\right)}^2]$

$=\frac{3n^2}{2x/6}=\frac{3n^2}{32}$

Comment: Correct answer is $C)\frac{3n^2}{2}$

but meashed answer is $a)0$ which is wrong