Integrate WITHOUT substitution rule.

Question

calculator · Answer

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

anonymous · Answer

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

calculator · Answer

not bad, how you think so fast?

calculator · Answer

@Ishaan94

anonymous · Answer

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 · Answer

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

dumbcow · Answer

whats wrong with substitution?

anonymous · Answer

Must be a homework question

anonymous · Answer

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 · Answer

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

dumbcow · Answer

u = x-2  --> x = u+2
du = dx

$$\int\limits_{?}^{?}\frac{u+2}{u} = \int\limits_{\int\limits_{?}^{?}}^{?}1 + \frac{2}{u} $$

anonymous · Answer

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