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