Integral using substitution/parts

[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 can't 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?

 

[jamesbob]

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

"By parts" instead of "Byparts" ?

 

[VietDao29]

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 can't 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 ?

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?

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?

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

 

[VietDao29]

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

"By parts" instead of "Byparts" ?

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.

 

[jamesbob]

Cheers!

When doing the polynomial division i got

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

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

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

 

[jamesbob]

wait if this is correct then i don't need partial fractions as

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

 

[jamesbob]

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

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

 

[VietDao29]

Cheers!

When doing the polynomial division i got

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

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

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

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.
----------------

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

Correct, $$\int\frac{dx}{1 + x^2} = \arctan x + C$$
Don't forget dx, and +C.
----------------

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

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. :)