Suppose I want to find the indefinite integral for $\frac {a(x)}{b(x)}$ where $a(x)=p(x)q(x)$, am I correct in solving it this way: $$\begin{align} \int \frac {a(x)}{b(x)} dx&=\frac1{b(x)}\int a(x) \ dx-\int\biggl(\frac1{b(x)}\biggr)'\biggl(\int a(x) \ dx\biggr) \ dx\\ &=\frac1{b(x)}\int p(x)q(x) \ dx-\int\biggl(\frac1{b(x)}\biggr)'\biggl(\int p(x)q(x) \ dx\biggr) \ dx\\ &=\frac1{b(x)}\biggl(p(x)\int q \ dx-\int p'(x)\Bigl(\int q(x) \ dx\Bigr) \ dx\biggr)-\int\biggl(\frac1{b(x)}\biggr)'\biggl(p(x)\int q(x) \ dx-\int p'(x)\Bigl(\int q(x) \ dx\Bigr) \ dx\biggr) \ dx \end{align} $$ Is this integral correct? Or are there principles I am overlooking? And also, are there other things that I should keep in mind?

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    $\begingroup$ If $\;b\;$ is in fact a non-constant function of $\;x\;$ then the first step is completely wrong: you can not take out of the integral such a term. $\endgroup$ – DonAntonio Mar 18 '18 at 12:45
  • $\begingroup$ @DonAntonio So if I understood correctly, $$\int\frac{2x}{3x^2}dx\neq\int\frac1{3x^2}\cdot2xdx$$ ?? $\endgroup$ – John Glenn Mar 18 '18 at 12:49
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    $\begingroup$ @JohnGlenn That is not what you did in your very first step... $\endgroup$ – DonAntonio Mar 18 '18 at 12:51
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    $\begingroup$ @DonAntonio Why is the first step wrong? He is applying integration by parts. $\endgroup$ – F.A. Mar 18 '18 at 13:02
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    $\begingroup$ @F.A. In fact, but the notation is quite misleading. $\endgroup$ – Peter Mar 18 '18 at 13:03

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