I'm trying to solve

$$\frac{(\sqrt x + x)\,dy}{dx} = \sqrt y + y$$

I can separate the variables and get

$$\frac {dy} {\sqrt y + y} = \frac{dx}{\sqrt x + x}$$

I know that integrating $$\frac {1}{x} = \ln (x)$$ but in this case why can't I just say that the integral of the above with reference to $x$ is $\ln$ (denominator)?

  • 1
    $\begingroup$ Change variables with $u=\sqrt{x}$, $v=\sqrt{y}$, and solve $$\frac{dv}{1+v}=\frac{du}{1+u}.$$ $\endgroup$ – Did Jan 30 '16 at 22:29
  • $\begingroup$ Use $\frac {dy} {\sqrt y + y}$ to show $\frac {dy} {\sqrt y + y}$, for example. Formatting tips here. $\endgroup$ – Em. Jan 30 '16 at 22:44
  • $\begingroup$ @Did, thank you. I would like to know the intuition behind why $ln (\sqrt(x)*x)$ is wrong if someone could clarify. $\endgroup$ – rddead Jan 30 '16 at 23:46
  • $\begingroup$ The derivative of $\log f(x)$ is $f'(x)/f(x)$ when $f$ is differentiable and $f(x)>0$. The derivative of $\log f(x)$ is not $1/f(x)$ except when $f(x)=x$, that is ,when $f'(x)=1$. So $\int (1/g(x) dx \ne \log g(x)$ except when $g(x)=x$ $\endgroup$ – DanielWainfleet Jan 31 '16 at 6:22
  • $\begingroup$ Sorry but I would be at a loss to describe or define "the intuition behind why ln((√x)∗x) is wrong". Where is ln((√x)∗x) coming from? Please explain what you mean. $\endgroup$ – Did Jan 31 '16 at 8:28


By changing the variable you get:

$$\int\frac{dx}{x+\sqrt{x}}\underbrace{=}_{x=t^2} \int\frac{2dt}{t+1}=2\ln |t+1|+c\underbrace{=}_{x=t^2}2\ln (\sqrt{x}+1) +c.$$


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