# Math Help - Integration problem

1. ## Integration problem

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

2. Can i split it up and do 1/(1+x^2) dx and arctanx dx? then multiply the results?

3. No there is no product rule or quotient rule for integration

Let u = arctan(x)

Then du = 1/(1+x^2) dx

int(arctan(x)/(1+x^2)dx) = int (udu) = u^2/2 +C = [arctan(x)^2]/2+C

4. Originally Posted by Calculus26
No there is no product rule or quotient rule for integration
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

5. 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

6. Originally Posted by Calculus26
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.
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.

7. hmm, still a bit unsure about this, ill try woking through it and post later to let you know how i got on, thanks everyone.

8. Originally Posted by Calculus26
No there is no product rule or quotient rule for integration

Let u = arctan(x)

Then du = 1/(1+x^2) dx

int(arctan(x)/(1+x^2)dx) = int (udu) = u^2/2 +C = [arctan(x)^2]/2+C
In this substitution why cant i take 1+x^2 as u?

9. Originally Posted by offahengaway and chips
In this substitution why cant i take 1+x^2 as u?
Exactly! That was what Calculus26 was saying. Unfortunately, tom then started talking about "integration by parts", confusing things.

10. so it doesnt matter what value i take as u? is that right?

11. 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.

12. Ok so

int (arctanx/1+x^2) dx

u = arctanx

du/dx = 1/1+x^2

so:

int(arctanx/1+x^2) dx = this is where i am stuck

13. Originally Posted by offahengaway and chips
Ok so

int (arctanx/1+x^2) dx

u = arctanx

du/dx = 1/1+x^2

so:

int(arctanx/1+x^2) dx = this is where i am stuck
$\int{\frac{\arctan{x}}{1 + x^2}\,dx} = \int{\arctan{x}\,\frac{1}{1 + x^2}\,dx}$.

Let $u = \arctan{x}$ so that $\frac{du}{dx} = \frac{1}{1 + x^2}$.

So $\int{\arctan{x}\,\frac{1}{1 + x^2}\,dx} = \int{u\,\frac{du}{dx}\,dx}$

$= \int{u\,du}$

$= \frac{1}{2}u^2 + C$

$= \frac{1}{2}(\arctan{x})^2 + C$.

14. 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)

15. so int udu= int(arctanx)du

what is du?

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