anonymous
  • anonymous
Indefinite Integral Help int (x/(2x+1) dx I keep getting (x/2)-((1/4)ln|2x+1|)+1/4+C Book says ((1/2)x)-((1/4)ln|2x+1|)+C. So when I get a constant as part of the answer when trying to find the indefinite integral, do I just consider it as part of the constant C? Is that why I'm off by +1/4?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
yes, 1/4 is absorbed into C, because C is unknown that will happen a lot in differential equations, so get used to it
Akshay_Budhkar
  • Akshay_Budhkar
yes whenever you have another constant, you ought to add it into C. The C absorbs it.. If you are having marks for steps then this step will count for marks as well ( maybe 1/2 mark or so if u r scoring full elsewhere)
TuringTest
  • TuringTest
wolfram keeps the 1 though ironically http://www.wolframalpha.com/input/?i=int%20(x%2F(2x%2B1)%20dx&t=crmtb01 but they left it in the parentheses, so I guess they feel it's justified

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Akshay_Budhkar
  • Akshay_Budhkar
but in the alternate forms it does absorb the constant :D
TuringTest
  • TuringTest
right^
Akshay_Budhkar
  • Akshay_Budhkar
turing i need u 1 min
Akshay_Budhkar
  • Akshay_Budhkar
i will give u the link
TuringTest
  • TuringTest
I don't even see why they add the 1 for 'restricted values of x' in wolf what does that mean, do you know akshay?
myininaya
  • myininaya
what are you guys doing?
myininaya
  • myininaya
comparing your answers with wolfram?
myininaya
  • myininaya
sprusty 1/4+constant is still a constant so your answer and the book answer are fine! :)
TuringTest
  • TuringTest
yes, but why does wolf add the 1 for 'restricted values' I want to know
myininaya
  • myininaya
\[\int\limits_{}^{}\frac{x}{2x+1} dx\] u=2x+1 => du=2 dx u=2x+1 => (u-1)/2=x \[\int\limits_{}^{}x \cdot \frac{1}{2x+1} dx=\int\limits_{}^{}\frac{u-1}{2} \cdot \frac{1}{u} \frac{du}{2}\] \[\frac{1}{4}\int\limits_{}^{}\frac{u-1}{u} du=\frac{1}{4}\int\limits_{}^{}(1-\frac{1}{u}) du=\frac{1}{4}(u-\ln|u|)+C\]
myininaya
  • myininaya
\[=\frac{1}{4}(2x+1-\ln|2x+1|)+C\]
myininaya
  • myininaya
\[=\frac{x}{2}+\frac{1}{4}-\frac{1}{4} \ln|2x+1| +C\]
myininaya
  • myininaya
\[=\frac{x}{2}-\frac{1}{4}\ln|2x+1|+c \]
myininaya
  • myininaya
i will look at wolfram's solution one sec
myininaya
  • myininaya
No I don't know what they are talking about honestly lol
myininaya
  • myininaya
either one of those things they have listed is fine though
myininaya
  • myininaya
imperialist seems smart I will ask him what he think wolfram means

Looking for something else?

Not the answer you are looking for? Search for more explanations.