Question

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

• A. $\ln x+\frac{x^3}{2}+c$
• B. $\ln x+3x^3+c$
• C. $\ln x+\frac{x^3}{3}+c$
• D. $\ln x+x^3+c$

Answer 1

The correct option is D $\ln x+x^3+c$

$\frac{dy}{dx}=\frac{1}{x}+3x^2$

$\Rightarrow dy=(\frac{1}{x}+3x^2)dx$

$\Rightarrow \int dy=\int (\frac{1}{x}+3x^2)dx$, integrate both sides

$\Rightarrow y=\ln x+3\cdot \frac{x^3}{3}+c=\ln x+x^3+c$