CBSE 2022 · Region 2 · Set 1 · Q8 · 3 marks
Evaluate : \[\int_{0}^{1} \tan ^{-1} x d x \]Find : \[\int \frac{2 x}{x^{2}+3 x+2} d x \]
Evaluate : \[\int_{0}^{1} \tan ^{-1} x d x \]
Find : \[\int \frac{2 x}{x^{2}+3 x+2} d x \]
Marking-scheme solution
(a)
Consider $\displaystyle \int (\tan^{-1} x)\,dx = \tan^{-1} x \cdot x - \int \dfrac{1}{1+x^{2}}\cdot x\,dx$
\[= x\,\tan^{-1}x - \tfrac{1}{2}\log(1+x^{2})\]
\[\int_{0}^{1}(\tan^{-1}x)\,dx = \left. x\,\tan^{-1}x - \tfrac{1}{2}\log(1+x^{2})\right]_{0}^{1}\]
\[= \dfrac{\pi}{4} - \dfrac{1}{2}\log 2\]
Or(b)\[I = \int \dfrac{2x\,dx}{x^{2}+3x+2} = \int \dfrac{2x}{(x+1)(x+2)}\,dx\]
\[= \int\left(\dfrac{-2}{x+1} + \dfrac{4}{x+2}\right)dx \qquad \text{using partial fraction}\]
\[= -2\log|x+1| + 4\log|x+2| + C\]
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.