CBSE 2025 · Region 1 · Set 3 · Q27 · 3 marks
Find : $\displaystyle \int \frac{2 x-1}{(x-1)(x+2)(x-3)} \mathrm{d} x$Evaluate : $\displaystyle \int_{0}^{5}(|x-1|+|x-2|+|x-5|) \mathrm{d} x$
Find : $\displaystyle \int \frac{2 x-1}{(x-1)(x+2)(x-3)} \mathrm{d} x$
Evaluate : $\displaystyle \int_{0}^{5}(|x-1|+|x-2|+|x-5|) \mathrm{d} x$
Marking-scheme solution
\[\begin{array}{r}
\int \frac{2 x-1}{(x-1)(x+2)(x-3)} \mathrm{d} x=-\frac{1}{6} \int \frac{1}{x-1} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x+2} \mathrm{d} x+\frac{1}{2} \int \frac{1}{x-3} \mathrm{d} x \\
\text { (Using Partial Fraction) } \\
=-\frac{1}{6} \log |x-1|-\frac{1}{3} \log |x+2|+\frac{1}{2} \log |x-3|+C
\end{array}
\]
\[\begin{aligned}
& I=\int_{0}^{5}(|x-1|+|x-2|+|x-5|) \mathrm{d} x \\
& \therefore I=\left[-\int_{0}^{1}(x-1) \mathrm{d} x+\int_{1}^{5}(x-1) \mathrm{d} x\right]+\left[-\int_{0}^{2}(x-2) \mathrm{d} x+\int_{2}^{5}(x-2) \mathrm{d} x\right]+\left[-\int_{0}^{5}(x-5) \mathrm{d} x\right] \\
& \left.\left.\left.\left.\left.=-\frac{(x-1)^{2}}{2}\right]_{0}^{1}+\frac{(x-1)^{2}}{2}\right]_{1}^{5}-\frac{(x-2)^{2}}{2}\right]_{0}^{2}+\frac{(x-2)^{2}}{2}\right]_{2}^{5}-\frac{(x-5)^{2}}{2}\right]_{0}^{5} \\
& =\frac{17}{2}+\frac{13}{2}+\frac{25}{2}=\frac{55}{2}
\end{aligned}
\]
IntegralsIntegration by Partial FractionsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.