CBSE 2026 · Region 1 · Set 1 · Q26 · 3 marks
Evaluate : $\displaystyle \int_{0}^{1} x \tan ^{-1} x \mathrm{~d} x$.
Official answer
From CBSE’s own marking scheme for this paper.
π/$\displaystyle 8$ - $\displaystyle 1$/$\displaystyle 4$
Marking-scheme solution
$\displaystyle \int_0^1 x\tan^{-1}x\,dx = \left[\tan^{-1}x \times \dfrac{x^2}{2}\right]_0^1 - \int_0^1 \dfrac{1}{1+x^2}\times\dfrac{x^2}{2}\,dx$
$\displaystyle = \dfrac{\pi}{8} - \int_0^1 \dfrac{1+x^2-1}{1+x^2}\times\dfrac{1}{2}\,dx$
$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2}\left[\int_0^1 dx - \int_0^1 \dfrac{1}{1+x^2}\,dx\right]$
$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2}\left[x-\tan^{-1}x\right]_0^1$
$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2} + \dfrac{\pi}{8} = \dfrac{\pi}{4} - \dfrac{1}{2}$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.