Why can't I use the quotient rule to take the integral of xdx/(x+1)? And what are the steps to solve that?

Answer 1

The quotient rule can not be used to solve this integral because I believe the quotient rule only applies to problems you are trying to take the derivative of.
The way I believe this problem needs to be done is to take the integral by substitution. Make u=(x+1), du would then equal 1dx. If you then solve for x, (u-1)=x. You could substitute this into the problem to make it ((u-1)du)/u.
Then split this the numerator up so that the problem looks like (u/u)-(1/u)du. If you reduce it equals 1-(1/u)du. Then taking the intergral would make it x-ln(absolute value of u). Substitue back in what u equals. Making the final answer, x-ln(x+1)+C.

Answer 2

Rule of thumb: When you have a fraction of polynomials, and the degree of the top is greater than or equal to the degree on the bottom, use long division to change the form of it.

In this case, the degree of the numerator is 1, and the degree of the denominator is 1. Since they're equal, use long division on (x+1) and x. I won't go over the details, but the answer should be

x/(x+1) = -1/(x+1) + 1

That is MUCH simpler to take the integral of.

Integral (x/(x+1))dx = Integral (-1/(x+1) + 1)dx

First, split it up into two integrals.

Integral (-1/(x+1) ) dx + Integral (1) dx

Pull out all constants.

(-1) Integral (1/(x+1))dx + Integral (1)dx

And now it's trivial to solve.

(-1) [ ln | x + 1 | ] + x + C

-ln|x + 1| + x + C

You can't use the quotient rule (or product rule) to solve for integrals because that's just not how things work out. The product and quotient rule were derived from the definition of the derivative (limit as h approches 0 of f(x+h) - f(x) over h). Integration is much more difficult than differentiation because it involves working _with_ the derivative laws in existence and finding techniques to work around it, and backwards.

Answer 3

As mentioned by the previous person, the quotient rule is used to find the derivative of a quotient.
The way I would approach the integral is by first dividing x/x+1. Notice, x/x+1=1-1/(x+1). We know that the intergral of 1 is x and the integral of 1/(x+1) is ln(x+1). As a result we get that the integral of x/x-6= x-ln(x+1)+C.