Mathematics

If $$\dfrac{dy}{dx}=\dfrac{1}{x}+3x^2$$ then $$y=$$


ANSWER

$$\ln{x} +x^3+c$$


SOLUTION
$$\dfrac{dy}{dx}=\dfrac{1}{x}+3x^2$$
$$\Rightarrow dy=\left(\dfrac{1}{x}+3x^2\right)dx$$
$$\Rightarrow \displaystyle \int dy=\int\left(\dfrac{1}{x}+3x^2\right)dx$$, integrate both sides 
$$\Rightarrow y=\ln x+3\cdot \dfrac{x^3}{3}+c=\ln x+x^3+c$$
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Single Correct Medium Published on 17th 09, 2020
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