Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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