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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 4 · Set 1 · Q26
Find: ∫ (x^2+ x +1)/(( x +1)^2( x +2)) d x
Find : \[\int \frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)} d \mathrm{x} \]
Marking-scheme solution
$$\begin{gathered}
\text { Let } \mathrm{I}=\int \frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)} d \mathrm{x} \\
\text { Here } \frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}+1)^{2}}+\frac{\mathrm{C}}{\mathrm{x}+2} \\
\Rightarrow \mathrm{x}^{2}+\mathrm{x}+1=\mathrm{A}(\mathrm{x}+1)(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+2)+\mathrm{C}(\mathrm{x}+1)^{2}
\end{gathered}
$$
On comparing, we get
$$\mathrm{A}=-2, \mathrm{B}=1 \text { and } \mathrm{C}=3 .
$$
$$\therefore \mathrm{I}=\int \frac{-2 \mathrm{dx}}{\mathrm{x}+1}+\int \frac{\mathrm{dx}}{(\mathrm{x}+1)^{2}}+3 \int \frac{\mathrm{dx}}{\mathrm{x}+2}
$$
$$=-2 \log |\mathrm{x}+1|-\frac{1}{\mathrm{x}+1}+3 \log |\mathrm{x}+2|+\mathrm{C}
$$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.