Indefinite Integral

• February 22nd 2010, 07:11 PM
Archduke01
Indefinite Integral
$\int (2/x^5 + 16 / 11x) dx$

I'm not sure what I'm doing wrong here. I end up with $2 \int 6/x^6 + 16 \int 2/11x^2$
• February 22nd 2010, 07:17 PM
pickslides
Quote:

Originally Posted by Archduke01
$\int (2/x^5 + 16 / 11x) dx$

I'm not sure what I'm doing wrong here. I end up with $2 \int 6/x^6 + 16 \int 2/11x^2$

try $
2 \int x^{-5}~dx+ \frac{16}{11}\int \frac{1}{x}~dx

$
• February 22nd 2010, 07:25 PM
Archduke01
$2 \int x^-4 /-4 + 16/11 \int x^-1 dx$

*the first integral is supposed to be x^-4 but it won't show up as a power when I put the math tags... sorry. I'm not sure what to do with the 2nd integral.
• February 22nd 2010, 07:27 PM
Prove It
Quote:

Originally Posted by Archduke01
$2 \int x^-4 /-4 + 16/11 \int x^-1 dx$

*the first integral is supposed to be x^-4 but it won't show up as a power when I put the math tags... sorry. I'm not sure what to do with the 2nd integral.

Hint: $\frac{d}{dx}(\ln{|x|}) = \frac{1}{x}$...
• February 22nd 2010, 07:32 PM
Archduke01
$2 \int x^-4/-4 + 16/11 \int ln x + C$

Would that be the final answer?

What is up with that x^-4?
• February 22nd 2010, 07:42 PM
Jhevon
Quote:

Originally Posted by Archduke01
$2 \int x^-4/-4 + 16/11 \int ln x + C$

Would that be the final answer?

What is up with that x^-4?

if your power has more than one "elements" in it, put them in {}-brackets. $$x^-4$$ gives $x^-4$ while $$x^{-4}$$ gives $x^{-4}$

notice that Prove It had $\ln |x|$ not $\ln x$

and when you take anti-derivative, you drop the $\int$ signs...
• February 22nd 2010, 07:44 PM
Archduke01
I love you Jhevon, thank you for your patience. I apologize for the stupid questions.