From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	By dividing the two terms in the numerator by <math>x</math>, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand,
-	<center> [[Image:2_1_3d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	\int \frac{x^{2}+1}{x}\,dx
		+	&= \int \Bigl(\frac{x^2}{x} + \frac{1}{x}\Bigr)\,dx\\[5pt]
		+	&= \int \bigl(x+x^{-1}\bigr)\,dx\\[5pt]
		+	&= \frac{x^2}{2} + \ln |x| + C\,,
		+	\end{align}</math>}}
		+
		+	where <math>C</math> is an arbitrary constant.
		+
		+
		+	Note: Observe that <math>1/x</math> has a singularity at <math>x=0</math>, so the answers above are only primitive functions over intervals that do not contain <math>x=0\,</math>.

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Current revision

By dividing the two terms in the numerator by \displaystyle x, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand,

\displaystyle \begin{align} \int \frac{x^{2}+1}{x}\,dx &= \int \Bigl(\frac{x^2}{x} + \frac{1}{x}\Bigr)\,dx\\[5pt] &= \int \bigl(x+x^{-1}\bigr)\,dx\\[5pt] &= \frac{x^2}{2} + \ln |x| + C\,, \end{align}

where \displaystyle C is an arbitrary constant.

Note: Observe that \displaystyle 1/x has a singularity at \displaystyle x=0, so the answers above are only primitive functions over intervals that do not contain \displaystyle x=0\,.