Lösung 2.1:5a
Aus Online Mathematik Brückenkurs 2
If we multiply top and bottom of the fraction by the conjugate expression \displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares gives that denominator's root is squared away,
| \displaystyle \begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}} &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} \end{align} |
Thus,
| \displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.} |
If we write the square roots in power form,
| \displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,, |
we see that we have a standard integral and can write down the primitive functions directly,
| \displaystyle \begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] &= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] &= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] &= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,, \end{align} |
where C is an arbitrary constant.
This can also be written with square roots as
| \displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.} |
Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
| \displaystyle \begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr) &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} \end{align} |
