From Förberedande kurs i matematik 2

Revision as of 09:30, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_3d.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:34, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Observe that the derivative of the denominator is, for the most part, equal to the numerator,
-	<center> [[Bild:2_2_3d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>(x^2+2x+2)' = 2x+2 = 2(x+1)</math>}}
		+
		+	so we can rewrite the integral as
		+
		+	{{Displayed math||<math>\int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}</math>}}
		+
		+	The substitution <math>u=x^2+2x+2</math> will therefore simplify the integral considerably,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int \frac{x+1}{x^2+2x+2}\,dx
		+	&= \left\{\begin{align}
		+	u &= x^2+2x+2\\[5pt]
		+	du &= (x^2+2x+2)'\,dx = 2(x+1)\,dx
		+	\end{align}\right\}\\[5pt]
		+	&= \frac{1}{2}\int \frac{du}{u}\\[5pt]
		+	&= \frac{1}{2}\ln |u| + C\\[5pt]
		+	&= \frac{1}{2}\ln |x^2+2x+2| + C\,\textrm{.}
		+	\end{align}</math>}}
		+
		+
		+	Note: By completing the square
		+
		+	{{Displayed math||<math>x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1</math>}}
		+
		+	we see that <math>x^2+2x+2</math> is always greater than or equal to 1, so we can take away the absolute sign around the argument in <math>\ln</math> and answer with
		+
		+	{{Displayed math||<math>\frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}</math>}}

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Current revision

Observe that the derivative of the denominator is, for the most part, equal to the numerator,

\displaystyle (x^2+2x+2)' = 2x+2 = 2(x+1)

so we can rewrite the integral as

\displaystyle \int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}

The substitution \displaystyle u=x^2+2x+2 will therefore simplify the integral considerably,

\displaystyle \begin{align} \int \frac{x+1}{x^2+2x+2}\,dx &= \left\{\begin{align} u &= x^2+2x+2\\[5pt] du &= (x^2+2x+2)'\,dx = 2(x+1)\,dx \end{align}\right\}\\[5pt] &= \frac{1}{2}\int \frac{du}{u}\\[5pt] &= \frac{1}{2}\ln |u| + C\\[5pt] &= \frac{1}{2}\ln |x^2+2x+2| + C\,\textrm{.} \end{align}

Note: By completing the square

\displaystyle x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1

we see that \displaystyle x^2+2x+2 is always greater than or equal to 1, so we can take away the absolute sign around the argument in \displaystyle \ln and answer with

\displaystyle \frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}