Hi Im having a bit of trouble getting started on the following integration problem,
any hints would be great
integral of arctanx/1+x^2 dx
Except that the product rule is exactly what you use, and work backwards through, whenever you do integration by parts. E.g., Balloon Calculus Forum - Viewing Thread
And the quotient rule is just a special case of the product rule. You might integrate the square of cosec x by working back through it. Balloon Calculus: standard integrals and derivatives
While I'm here, here's (or will be in two tics) a picture for the present (chain rule) problem, just in case it helps...
Don't integrate - balloontegrate
Balloon Calculus Forum
I agree that int by parts is a product rule of sorts for certain types of products there is no product rule that works on every product as with differentiation.
In the same light partial fraction decompostion is a quotient rule if you will for rational functions but again there is no integration rule for quotients in general as with differentiation
Point taken. A product is just as likely to be (as in the present problem) the result of a chain-rule differentiation as a product-rule one. There's no version of either kind of rule for integration, though integration is nearly always a matter of working back through those two rules for differentiation.
No No No
Lets look at this again
when you take u = arctan(x)
du = 1/(1+x^2)dx which yields (1+x^2)du = dx
int(arctan(x)/(1+x^2)dx = int (u (1+x^2)/(1+x^2)dx) = int(udu) which can be easily integrated
If you let u = 1 + x^2 then du = 2x dx
you get int (arctan(x)/(2xu)du) which is a mess.
The idea of u substitution is to obtain an integral in terms of u which you can integrate.
In general there is only one choice maybe two which will accomplish this.
What substitution to make ? thatg comes from experience , insight, and good ol' trial and error.
$\displaystyle \int{\frac{\arctan{x}}{1 + x^2}\,dx} = \int{\arctan{x}\,\frac{1}{1 + x^2}\,dx}$.
Let $\displaystyle u = \arctan{x}$ so that $\displaystyle \frac{du}{dx} = \frac{1}{1 + x^2}$.
So $\displaystyle \int{\arctan{x}\,\frac{1}{1 + x^2}\,dx} = \int{u\,\frac{du}{dx}\,dx}$
$\displaystyle = \int{u\,du}$
$\displaystyle = \frac{1}{2}u^2 + C$
$\displaystyle = \frac{1}{2}(\arctan{x})^2 + C$.
Ok the method is called substitution because we substitute
we have int (arcatan(x)/(1+x^2)dx
We'll go one step at a time
Since u = arctan(x) substitute arctan(x) with u
int (arcatan(x)/(1+x^2)dx = int(u/(1+x^2)dx)
Now dx = (1+x^2)du so replace dx with this result
int(u/(1+x^2)dx) = int(u/(1+x^2) *(1+x^2)du)
Note (1+x^2)/(1+x^2) cancels leaving int(udu)