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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 2 · Set 2 · Q33
Find: ∫ (5 x)/(( x +1)( x^2+9)) d x.
Find : $\displaystyle \int \frac{5 \mathrm{x}}{(\mathrm{x}+1)\left(\mathrm{x}^{2}+9\right)} \mathrm{d} \mathrm{x}$.
Marking-scheme solution
\[\begin{aligned}
& \frac{5 \mathrm{x}}{(\mathrm{x}+1)\left(\mathrm{x}^{2}+9\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+9} \\
& \Rightarrow \mathrm{~A}=-\frac{1}{2}, \mathrm{~B}=\frac{1}{2}, \mathrm{C}=\frac{9}{2}
\end{aligned}
\]
Given integral
\[\begin{aligned}
& =-\frac{1}{2} \int \frac{1}{\mathrm{x}+1} \mathrm{d} \mathrm{x}+\frac{1}{2} \int \frac{\mathrm{x}+9}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x} \\
& =-\frac{1}{2} \int \frac{1}{\mathrm{x}+1} \mathrm{d} \mathrm{x}+\frac{1}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x}+\frac{1}{4} \int \frac{18}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x} \\
& =-\frac{1}{2} \log |\mathrm{x}+1|+\frac{1}{4} \log \left(\mathrm{x}^{2}+9\right)+\frac{3}{2} \tan ^{-1} \frac{\mathrm{x}}{3}+\mathrm{C}
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.