Question

Integrate WITHOUT substitution rule.

Answer 1

\[\int\limits_{}^{}\frac{x}{x-2}~dx\]

Answer 2

\[\frac{x -2 + 2}{x-2} = 1 + \frac{2}{x-2}\]I hope you can do it from here.

Answer 3

not bad, how you think so fast?

Answer 4

@Ishaan94

Answer 5

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.

Answer 6

This sort of heuristic reasoning only comes with practice, though.

Answer 7

whats wrong with substitution?

Answer 8

Must be a homework question

Answer 9

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 \(\frac{dg}{dx}\cdot f\neq Cx^0\).

Answer 10

It will, but I won't recommend substitution for such problems.

Answer 11

u = x-2 --> x = u+2
du = dx

\[\int\limits_{?}^{?}\frac{u+2}{u} = \int\limits_{\int\limits_{?}^{?}}^{?}1 + \frac{2}{u} \]

Answer 12

Partial fractions first in that example; you can't use partial faction decomposition for any arbitrary function, only for first degree ones.