# Thread: Indefinite Integral

1. ## Indefinite Integral

$\displaystyle \int (2/x^5 + 16 / 11x) dx$

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

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

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

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

3. $\displaystyle 2 \int x^-4 /-4 + 16/11 \int x^-1 dx$

*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.

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

*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: $\displaystyle \frac{d}{dx}(\ln{|x|}) = \frac{1}{x}$...

5. $\displaystyle 2 \int x^-4/-4 + 16/11 \int ln x + C$

Would that be the final answer?

What is up with that x^-4?

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

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 $\displaystyle x^-4$ while $$x^{-4}$$ gives $\displaystyle x^{-4}$

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

and when you take anti-derivative, you drop the $\displaystyle \int$ signs...

7. I love you Jhevon, thank you for your patience. I apologize for the stupid questions.