Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन4 Marks
Question
सिद्ध कीजिए: $\int_{1}^{3} \frac{1}{x^{2}(x+1)} d x$ $=\frac{2}{3}+\log \frac{2}{3}$
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Answer
$\int_{1}^{3} \frac{1}{x^{2}(x+1)} d x$ $=\frac{2}{3}+\log \frac{2}{3}$ आंशिक भिन्न विधि के प्रयोग से, माना $\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$ ...(i) $\Rightarrow$ 1 = Ax(x + 1) + B(x + 1) + Cx2 ...(ii) x = 0, -1 समी (ii) में रखने पर, 1 = B(0 + 1) और 1 = C(1)2 $\Rightarrow$ B = 1 और C = 1 समी (ii) के दोनों पक्षों में x2 के गुणांकों की तुलना करने पर, 0 = A + C $\Rightarrow$ A = -C = -1 $\therefore$ $\int_{1}^{3} \frac{1}{x^{2}(x+1)} d x$ $=\int_{1}^{3}\left(-\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}\right) d x$ =[−log |x| − 1x + log |x + 1$]_{1}^{3}$ $=\left[\log \left|\frac{x+1}{x}\right|-\frac{1}{x}\right]_{1}^{3}$ $=\left(\log \left|\frac{4}{3}\right|-\frac{1}{3}\right)-\left(\log \left|\frac{2}{1}\right|-\frac{1}{1}\right)$ $=\left(\log \frac{4}{3}-\log 2\right)+\left(1-\frac{1}{3}\right)$ [$\because$ log$(\frac{m}{n})$ = log m - log n] $=\log \left(\frac{4}{3} \times \frac{1}{2}\right)+\frac{2}{3}$ $=\log \frac{2}{3}+\frac{2}{3}$
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