গণিত প্রশ্ন

AI-তৈরি নমুনা উত্তর

ধরি,
$I = \int \frac{e^{m \tan^{-1}x}}{(1 + x^2)^2} dx$

মনে করি, $\tan^{-1}x = \theta \implies x = \tan \theta$
অতএব, $dx = \sec^2 \theta \, d\theta$

সুতরাং,
$I = \int \frac{e^{m\theta}}{(1 + \tan^2 \theta)^2} \sec^2 \theta \, d\theta$
$= \int \frac{e^{m\theta}}{\sec^4 \theta} \sec^2 \theta \, d\theta = \int e^{m\theta} \cos^2 \theta \, d\theta$
$= \frac{1}{2} \int e^{m\theta} (1 + \cos 2\theta) d\theta$
$= \frac{1}{2} \int e^{m\theta} d\theta + \frac{1}{2} \int e^{m\theta} \cos 2\theta \, d\theta$

আমরা জানি, $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2}(a \cos bx + b \sin bx)$

অতএব,
$I = \frac{1}{2} \cdot \frac{e^{m\theta}}{m} + \frac{1}{2} \cdot \frac{e^{m\theta}}{m^2 + 4} (m \cos 2\theta + 2 \sin 2\theta) + C$

এখন, $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - x^2}{1 + x^2}$ এবং $\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} = \frac{2x}{1 + x^2}$

অতএব,
$I = \frac{e^{m \tan^{-1}x}}{2m} + \frac{e^{m \tan^{-1}x}}{2(m^2 + 4)} \left[ m \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \left( \frac{2x}{1 + x^2} \right) \right] + C$
$= \frac{e^{m \tan^{-1}x}}{2} \left[ \frac{1}{m} + \frac{m(1 - x^2) + 4x}{(m^2 + 4)(1 + x^2)} \right] + C$

উত্তর: $\frac{e^{m \tan^{-1}x}}{2} \left[ \frac{1}{m} + \frac{m(1 - x^2) + 4x}{(m^2 + 4)(1 + x^2)} \right] + C$ (যেখানে $C$ একটি সমাকলন ধ্রুবক)।