# Integral using substitution/parts

1. Feb 7, 2006

### jamesbob

We've been focusing on integration by parts or by substitution and i can't figure some out:

$$\int 4x^3\tan^-1xdx$$ Here $$\tan^-1x$$ means inverse tanx. I cant get the latex right for it.

my working so far:

$$By parts, u = \tan^-1x , \frac{dv}{dx} = 4x^3, \frac{du}{dx} = \frac{1}{1 + x^2}, v = \frac{4x^4}{4} = x^4 \Rightarrow x^4\tan^-1 - \int\frac{x^4}{1 + x^2}$$

It's the Integral $$\int\frac{x^4}{1 + x^2}$$ that i can't do ???

The one i'm completely stuck on how to even start is:

Find the general solution x(t) of the differential equation

$$\frac{dx}{dt} = \frac{t^2 + 3t + 5}{t^2 + 6t +34}$$

and the particular solution for which x(-3) = 0.

Can anyone help?

Last edited: Feb 7, 2006
2. Feb 7, 2006

### jamesbob

PS. how do you leave spaces in Latex, ie.

"By parts" instead of "Byparts" ?

3. Feb 7, 2006

### VietDao29

You may wat to try some Polynomial long division.
Remember that, if you are about to integrate something like:
$$\int \frac{P_1(x)}{P_2(x)} dx$$, where P1(x), and P2(x) are polynomials.
If the degree of the numerator (i.e P1(x)) is lower than the degree of the denominator (i.e P2(x)), you can use partial fraction, and then integrate it.
If the degree of the numerator (i.e P1(x)) is higher than or equal to the degree of the denominator (i.e P2(x)), you should try Polynomial long division first to split it to:
$$\int \frac{P_1(x)}{P_2(x)} dx = \int \left( Q(x) + \frac{P_3(x)}{P_2(x)} \right) dx$$, Now the degree of P3(x) is clearly lower than the degree of P2(x). You then you can try to Partial fraction it, and then integrate that.
Can you go from here?
$$\frac{dx}{dt} = \frac{t^2 + 3t + 5}{t^2 + 6t +34}$$
$$\Rightarrow dx = \frac{t^2 + 3t + 5}{t^2 + 6t +34} dt$$
Now integrate both sides gives:
$$\int dx = \int \frac{t^2 + 3t + 5}{t^2 + 6t +34} dt$$
$$x = \int \frac{t^2 + 3t + 5}{t^2 + 6t +34} dt$$.
Don't forget the constant of integration, C.
After integrating that, you can use the fact that x(-3) = 0 to find C.
Can you go from here? :)

Last edited: Feb 7, 2006
4. Feb 7, 2006

### VietDao29

You may want to try something like:
\mbox{blah... blah... blah...}
or:
\textit{blah... blah... blah...}
Example:
1. $$\mbox{By parts}$$
2. $$\textit{By parts}$$
3. $$By parts$$
You can click on any LaTeX image to see its code.
\mbox{blah... blah... blah...} will display exactly what you want in normal type, while \textit{blah... blah... blah...} will display exactly what you want in italics.
If you don't use any of them, then it will display everything in italics but spaces.
You may want to have a glance here:
Introducing LaTeX Math Typesetting (it's in Math & Science Tutorials board, the first board from the top). There are 3 guides in PDF. Let's have a look at them.

5. Feb 7, 2006

### jamesbob

Cheers!

When doing the polynomial division i got

$$(x^2 - 1) + \frac{1}{x^2 + 1}$$

Is this correct. If so, whats the rule for applying partial fractions to an irreducible again? i.e. we have

$$x^2 + 1$$ so $$x^2 = -1$$

6. Feb 7, 2006

### jamesbob

wait if this is correct then i dont need partial fractions as

$$\int\frac{1}{1 + x^2} = tan^{-1}x$$ right?

7. Feb 7, 2006

### jamesbob

Is it possible to integrate the function? If not, how do you plug the x(-3) in?

8. Feb 8, 2006

### VietDao29

No, you cannot use partial fraction on that.
Your denominator is (x2 + 1), it cannot be factored any further.
For short, if your denominator is a product of some polynomials of degree 2 or 1, you can use partial fraction. Note that every polynimial can be factored to be the product of some polynomials of degree less than or equal to 2.
----------------
Correct, $$\int\frac{dx}{1 + x^2} = \arctan x + C$$
Don't forget dx, and +C.
----------------
Yes, it is. It's possible to integrate that. Now the degree of the numerator is equal to the degree of the denominator, what should you do first? (Hint: Polynomial division).
From there, I think you can solve the problem.
I'll give you an example:
-----------------
Example 1:
$$\int \frac{dx}{x ^ 2 + 2x + 3} = \int \frac{dx}{(x + 1) ^ 2 + 2}$$
Use u-substitution:
u = x + 1 => du = dx
We have:
$$\int \frac{dx}{(x + 1) ^ 2 + 2} = \int \frac{du}{u ^ 2 + (\sqrt{2}) ^ 2} = \frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) + C$$. Change u back to x, we have:
$$\int \frac{dx}{x ^ 2 + 2x + 3} = \frac{1}{\sqrt{2}} \arctan \left( \frac{x + 1}{\sqrt{2}} \right) + C$$
---------------
Example 2:
$$\int \frac{x}{x ^ 2 + x + 3} dx$$
Take the derivative of the denominator with respect to x, we have:
(x2 + x + 3)' = 2x + 1.
Now:
$$\int \frac{x}{x ^ 2 + x + 3} dx = \frac{1}{2} \int \frac{2x}{x ^ 2 + x + 3} dx = \frac{1}{2} \int \frac{2x + 1 - 1}{x ^ 2 + x + 3} dx$$
$$= \frac{1}{2} \int \frac{2x + 1}{x ^ 2 + x + 3} dx - \frac{1}{2} \int \frac{1}{x ^ 2 + x + 3} dx$$
To integrate this:
$$\int \frac{2x + 1}{x ^ 2 + x + 3} dx$$
Let u = x2 + x + 3, that means du = 2x + 1 dx, so we have:
$$\int \frac{du}{u} = \ln |u| + C = \ln|x ^ 2 + x + 3| + C = \ln(x ^ 2 + x + 3) + C$$, we can drop the absolute value because: x2 + x + 3 > 0, for all x.
To integrate:
$$\int \frac{1}{x ^ 2 + x + 3} dx = \int \frac{dx}{\left(x + \frac{1}{2} \right) ^ 2 + \frac{11}{4}}$$. Now do as example 1, let u = x + 1 / 2, we have:
$$\int \frac{1}{x ^ 2 + x + 3} dx = \frac{2}{\sqrt{11}} \arctan \left( 2 \times \frac{u}{\sqrt{11}} \right) + C = \frac{2}{\sqrt{11}} \arctan \left( 2 \times \frac{x + \frac{1}{2}}{\sqrt{11}} \right)$$
$$= \frac{2}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C$$.
So we have:
$$\int \frac{x}{x ^ 2 + x + 3} dx = \frac{1}{2} \ln(x ^ 2 + x + 3) - \frac{1}{2} \times \frac{2}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C$$
$$= \frac{1}{2} \ln(x ^ 2 + x + 3) - \frac{1}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C$$
Now let's see if you can work out that integral. :)

Last edited: Feb 8, 2006