Calculator
  • Calculator
Integrate WITHOUT substitution rule.
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!
Calculator
  • Calculator
\[\int\limits_{}^{}\frac{x}{x-2}~dx\]
anonymous
  • anonymous
\[\frac{x -2 + 2}{x-2} = 1 + \frac{2}{x-2}\]I hope you can do it from here.
Calculator
  • Calculator
not bad, how you think so fast?

Looking for something else?

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

More answers

Calculator
  • Calculator
@Ishaan94
anonymous
  • anonymous
Speaking for myself, the solution was easy to see because we can eliminate partial and substitution (neither being reasonable methods). Also, since the nominator and denominator were of the same polynomial degree, I reasoned that partial fractions were the best way to solve the problem.
anonymous
  • anonymous
This sort of heuristic reasoning only comes with practice, though.
dumbcow
  • dumbcow
whats wrong with substitution?
anonymous
  • anonymous
Must be a homework question
anonymous
  • anonymous
Well, substitution literally would not work.\[\int\frac{f}{g}\,dx\]Where \(f,g=ax^n+bx^{n-1}\ldots+C\) has it so that \(\frac{dg}{dx}\cdot f\neq Cx^0\).
anonymous
  • anonymous
It will, but I won't recommend substitution for such problems.
dumbcow
  • dumbcow
u = x-2 --> x = u+2 du = dx \[\int\limits_{?}^{?}\frac{u+2}{u} = \int\limits_{\int\limits_{?}^{?}}^{?}1 + \frac{2}{u} \]
anonymous
  • anonymous
Partial fractions first in that example; you can't use partial faction decomposition for any arbitrary function, only for first degree ones.

Looking for something else?

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