Lösung 2.1:5a

Aus Online Mathematik Brückenkurs 2

Version vom 10:19, 11. Mär. 2009

If we multiply top and bottom of the fraction by the conjugate expression x+9+x  then the formula for the difference of two squares gives that denominator's root is squared away,

1x+9−x=1x+9−xx+9+xx+9+x=x+9+xx+92−x2=x+9−xx+9+x=9x+9+x.

Thus,

dxx+9−x=91x+9+xdx.

If we write the square roots in power form,

91(x+9)12+x12dx

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

91(x+9)12+x12dx=9121+1(x+9)12+1+21+1x12+1+C=9132(x+9)32+32x32+C=9132(x+9)32+32x32+C=227(x+9)32+227x32+C

where C is an arbitrary constant.

This can also be written with square roots as

227(x+9)x+9+227xx+C.

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

ddx227(x+9)32+227x32+C=22723(x+9)32−1+22723x32−1+0=91(x+9)12+91x12.