If Du is Smaller Then Dx

[JeffM]

Ok, I understand most of it, but how did the 1 (on the left) on \(\displaystyle 1 - \dfrac{1}{1 + x}\) disappear, and why is there a negative sign next to the ln at some point in the equation?

The minus sign below came from your long division

\(\displaystyle 2 * \int \left(1\ MINUS\ \dfrac{1}{x + 1}\right)\ dx = 2 * \left\{\int 1\ dx\ MINUS\ \int \dfrac{1}{x + 1}\ dx\right\} =\) Make sense so far?

\(\displaystyle 2 * \left\{1 * \int dx\ MINUS\ \int \dfrac{1}{x + 1}\ dx\right\} = \) A constant factor within an integral can be taken outside the integral.

\(\displaystyle 2 * \left\{\int dx\ MINUS \int \dfrac{1}{x + 1}\ dx\right\} =\) One times something is the same thing.

\(\displaystyle 2 * \left\{x + M\ MINUS \int \dfrac{1}{x + 1}\right\} =\) The integral of dx is x + a constant

\(\displaystyle 2 * \{x + M\ MINUS\ (ln|x + 1| + L)\} =\) Integral of 1 / (x + 1) dx is ln|x + 1| + a constant

\(\displaystyle 2 * (x + M\ MINUS\ ln|x + 1| - L) = 2 * (x + M\ MINUS\ ln|x + 1| + K) = \)

\(\displaystyle 2x - 2ln|x + 1| + C.\)

 

[Jason76]

The minus sign below came from your long division

\(\displaystyle 2 * \int \left(1\ MINUS\ \dfrac{1}{x + 1}\right)\ dx = 2 * \left\{\int 1\ dx\ MINUS\ \int \dfrac{1}{x + 1}\ dx\right\} =\) Make sense so far?

\(\displaystyle 2 * \left\{1 * \int dx\ MINUS\ \int \dfrac{1}{x + 1}\ dx\right\} = \) A constant factor within an integral can be taken outside the integral.

\(\displaystyle 2 * \left\{\int dx\ MINUS \int \dfrac{1}{x + 1}\ dx\right\} =\) One times something is the same thing.

\(\displaystyle 2 * \left\{x + M\ MINUS \int \dfrac{1}{x + 1}\right\} =\) The integral of dx is x + a constant

\(\displaystyle 2 * \{x + M\ MINUS\ (ln|x + 1| + L)\} =\) Integral of 1 / (x + 1) dx is ln|x + 1| + a constant

\(\displaystyle 2 * (x + M\ MINUS\ ln|x + 1| - L) = 2 * (x + M\ MINUS\ ln|x + 1| + K) = \)

\(\displaystyle 2x - 2ln|x + 1| + C.\)

Thanks Jeff. Got it now. But one last question. You said when constants were integrated (in this situation) they were pulled out. However, doesn't an integrated constant = the constant(x) ? For instance \(\displaystyle \int 5 dx = 5x + C\)

 

[JeffM]

Thanks Jeff. Got it now. But one last question. You said when constants were integrated (in this situation) they were pulled out. However, doesn't an integrated constant = the constant(x) ? For instance \(\displaystyle \int 5 = 5x + C\)

It does indeed, but you can view that as

\(\displaystyle \int 5\ dx = 5 * \int dx = 5(x + K) = 5x + 5K = 5x + C,\ where\ 5K = C.\)

With indefinite integrals, the added constant can be anything so you can add them, multiply them, square them, and you still end up with a constant (a different constant, undoubtedly, but something that is also undoubtedly independent of x.)

The basic point is that you want to learn some basic integrals, and if you are worrying about constant factors, you may not see the basic integral. So I tend to get them out of the integral right away. It's not absolutely necessary, but it frequently simplifies the task of recognizing basic integrals.

 

[Jason76]

It does indeed, but you can view that as\(\displaystyle \int 5\ dx = 5 * \int dx = 5(x + K) = 5x + 5K = 5x + C,\ where\ 5K = C.\)With indefinite integrals, the added constant can be anything so you can add them, multiply them, square them, and you still end up with a constant (a different constant, undoubtedly, but something that is also undoubtedly independent of x.)The basic point is that you want to learn some basic integrals, and if you are worrying about constant factors, you may not see the basic integral. So I tend to get them out of the integral right away. It's not absolutely necessary, but it frequently simplifies the task of recognizing basic integrals.

Actually, the \(\displaystyle (x + M)\) is really \(\displaystyle (x + C)\), then when you do the distributive property \(\displaystyle (2C)\) become\(\displaystyle 2(1)\) and so you get \(\displaystyle 2x + 2\) (and then \(\displaystyle 2x - 2\) and after you move the negative sign in front of \(\displaystyle \ln \)