Solution 2.1:3c
From Förberedande kurs i matematik 2
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| - | {{ | + | If we multiply the factors in the integrand together and use the power laws, |
| - | + | ||
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \int e^{2x}\bigl(e^x+1\bigr)\,dx | ||
| + | &= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt] | ||
| + | &= \int\bigl(e^{2x+x} + e^{2x}\bigr)\,dx\\[5pt] | ||
| + | &= \int{\bigl(e^{3x} + e^{2x}\bigr)}\,dx\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | we obtain a standard integral with two terms of the type <math>e^{ax}</math>, where | ||
| + | <math>a</math> is a constant. The indefinite integral is therefore | ||
| + | |||
| + | {{Displayed math||<math>\int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,,</math>}} | ||
| + | |||
| + | where <math>C</math> is an arbitrary constant. | ||
Current revision
If we multiply the factors in the integrand together and use the power laws,
| \displaystyle \begin{align}
\int e^{2x}\bigl(e^x+1\bigr)\,dx &= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt] &= \int\bigl(e^{2x+x} + e^{2x}\bigr)\,dx\\[5pt] &= \int{\bigl(e^{3x} + e^{2x}\bigr)}\,dx\,, \end{align} |
we obtain a standard integral with two terms of the type \displaystyle e^{ax}, where \displaystyle a is a constant. The indefinite integral is therefore
| \displaystyle \int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,, |
where \displaystyle C is an arbitrary constant.
