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प्रश्न
`(2ax+x^2)(dy)/(dx)=a^2+2ax`
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उत्तर
\[\left( 2ax + x^2 \right)\frac{dy}{dx} = a^2 + 2ax \]
\[\frac{dy}{dx} = \frac{a^2 + 2ax}{2ax + x^2} = \frac{a\left( a + 2x \right)}{x\left( 2a + x \right)} \]
\[\text{Let }x = 2a \tan^2 \theta \Rightarrow dx = 4a \tan\theta \sec^2 \theta\ d \theta \]
\[\frac{dy}{dx} = \frac{a\left( a + 4a\ tan^2 \theta \right)}{2a \tan^2 \theta \left( 2a \right)\left( 1 + \tan^2 \theta \right)}\]
\[\int dy = \int\frac{a \left( 1 + 4 \tan^2 \theta \right)}{2 \tan^2 \theta \left( 2a \right) \left( \sec^2 \theta \right)}dx \]
\[\int dy = \int\frac{a\left( 1 + 4 \tan^2 \theta \right)}{2 \tan^2 \theta \left( 2a \right) \left( \sec^2 \theta \right)}\left( 4a \right)\tan\theta \sec^2 \theta\ d\theta \]
\[= \int\frac{a \left( 1 + 4 \tan^2 \theta \right)}{\tan\theta}d\theta \]
\[= a\int\left( \frac{1}{\tan\theta} + 4\tan\theta \right)d\theta \]
\[y = a\int\cot\theta + 4\ tan\theta\ d\theta \]
\[ y = a\left[ \log \sin\theta + 4 \left( - \log \cos\theta \right) \right] + c \]
\[ y = a\left[ \log\sin\theta - 4\log \cos\theta \right] + c \]
\[\text{As, }x = 2a \tan^2 \theta \Rightarrow \tan\theta = \sqrt{\frac{x}{2a}} \]
\[y = a \log \left( \frac{\sin\theta}{\cos^4 \theta} \right) + c \]
\[= a\log\left( \frac{\tan\theta}{\cos^3 \theta} \right) + c \]
\[= a\log \left( \sqrt{\frac{x}{2a}} \times \left( \sqrt{\frac{x + 2a}{2a}} \right)^3 \right) + c \]
\[y = a\log\left( \frac{x^\frac{1}{2} {(x + 2a)}^\frac{3}{2}}{4 a^2} \right) + c\]
\[y + C = \frac{a}{2} \left( \log x + 3\log\left( x + 2a \right) \right)\text{ where }C = c - a\log\left( 4 a^2 \right)\]
