Indefinite Integral

brandy

New member
Joined
Jan 15, 2009
Messages
15
When finding the indefinite integral I don't understand why one method I tried works but the other doesn't. Can you help explain why the one that I marked as incorrect doesn't work.

CORRECT!

\(\displaystyle \int \, \left(x^2\, +\, 5\right)^3\, dx\)

\(\displaystyle =\, \int\, \left(x^6\, +\, 15x^4\, +\, 75x^2\, +\, 125\right)\, dx\)

\(\displaystyle =\, \int \, x^6\, dx\, +\, \int\, 15x^4\, dx\, +\, \int\, 75x^2\, dx\, +\, \int\, 125\, dx\)

\(\displaystyle =\, \frac{1}{7}x^7\, +\, 3x^5\, +\, 25x^3\, +\, 125x\, +\, C\)

I don't understand why the following doesn't work also???

INCORRECT???

\(\displaystyle \int\, \left(x^2\, +\, 5\right)^3\, dx\)

\(\displaystyle =\, \left(\frac{1}{2x}\right)\left(\frac{1}{4}\right)\left(x^2\, +\, 5\right)^4\, +\, C\)

\(\displaystyle = \left(\frac{1}{8x}\right)\left(x^2\, +\, 5\right)^4\, +\, C\)

\(\displaystyle \mbox{when expanded out this becomes}\)

\(\displaystyle \left(\frac{1}{8x}\right)\left(x^8\, +\, 20x^6\, +\, 50x^4\, +\, 500x^2\, +\, 625\right)\, +\, C\)

\(\displaystyle =\, \frac{1}{8}x^7\, +\, \frac{5}{2}x^5\, +\, \frac{75}{4}x^3\, +\, \frac{125}{2}x\, +\, \frac{625}{8x}\, +\, C\)
 
brandy said:
When finding the indefinite integral I don't understand why one method I tried works but the other doesn't. Can you help explain why the one that I marked as incorrect doesn't work. [attachment=0:2u9po2jq]Indefinite Integral.jpg[/attachment:2u9po2jq]


In the second method, you can't integrate (x^2 + 5)^3 dx, because you have u^3 but not du. You can try substituting y = x^2 + 5, but du = 2x dx, and there is no x "hanging around" to be used.
 
Top