Solution 2.2:3e
From Förberedande kurs i matematik 2
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| - | {{ | + | If we differentiate the denominator in the integrand |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>(x^2+1)' = 2x</math>}} |
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| + | we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as | ||
| + | |||
| + | {{Displayed math||<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}} | ||
| + | |||
| + | so the integral can be written as | ||
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| + | {{Displayed math||<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}} | ||
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| + | and we see that the substitution <math>u=x^2+1</math> can be used to simplify the integral, | ||
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| + | {{Displayed math||<math>\begin{align} | ||
| + | \int \frac{3x}{x^2+1}\,dx | ||
| + | &= \left\{ \begin{align} | ||
| + | u &= x^2+1\\[5pt] | ||
| + | du &= (x^2+1)'\,dx = 2x\,dx | ||
| + | \end{align}\right\}\\[5pt] | ||
| + | &= \frac{3}{2}\int \frac{du}{u}\\[5pt] | ||
| + | &= \frac{3}{2}\ln |u|+C\\[5pt] | ||
| + | &= \frac{3}{2}\ln |x^{2}+1|+C\\[5pt] | ||
| + | &= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | In the last step, we take away the absolute sign around the argument in <math>\ln</math>, because <math>x^2+1</math> is always greater than or equal to 1. | ||
Current revision
If we differentiate the denominator in the integrand
| \displaystyle (x^2+1)' = 2x |
we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as
| \displaystyle 3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)', |
so the integral can be written as
| \displaystyle \int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,, |
and we see that the substitution \displaystyle u=x^2+1 can be used to simplify the integral,
| \displaystyle \begin{align}
\int \frac{3x}{x^2+1}\,dx &= \left\{ \begin{align} u &= x^2+1\\[5pt] du &= (x^2+1)'\,dx = 2x\,dx \end{align}\right\}\\[5pt] &= \frac{3}{2}\int \frac{du}{u}\\[5pt] &= \frac{3}{2}\ln |u|+C\\[5pt] &= \frac{3}{2}\ln |x^{2}+1|+C\\[5pt] &= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.} \end{align} |
In the last step, we take away the absolute sign around the argument in \displaystyle \ln, because \displaystyle x^2+1 is always greater than or equal to 1.
