Aus Online Mathematik Brückenkurs 2

(HINT: multiply the top and bottom by the conjugate of the denominator.)

If we multiply top and bottom of the fraction by the conjugate expression, \displaystyle \sqrt{x+9}+\sqrt{x}, then the conjugate rule 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}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\ \end{align}

Thus,

\displaystyle \int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\left( \sqrt{x+9}+\sqrt{x} \right)}\,dx

If we write the square roots in power form,

\displaystyle \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx,

we see that we have a standard integral and can write down the primitive functions directly:

\displaystyle \begin{align} & \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx=\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right)+C \\ & =\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right)+C \\ & =\frac{1}{9}\left( \frac{2}{3}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{3}{2}} \right)+C \\ & =\frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \\ & \\ \end{align},

where C is an arbitrary constant.

This can also be written with square roots as

\displaystyle \frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C

To be completely certain that we have everything correctly, we differentiate the answer and see if we get back the integrand:

\displaystyle \begin{align} & \frac{d}{dx}\left( \frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \right) \\ & =\frac{2}{27}\centerdot \frac{3}{2}\left( x+9 \right)^{\frac{3}{2}-1}+\frac{2}{27}\centerdot \frac{3}{2}x^{\frac{3}{2}-1}+0 \\ & =\frac{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\ \end{align}