# Three integrals

## Homework Statement

$$1)\ \int \frac{1}{(x^2+4)^3}\ \mbox{d}x$$

$$2)\ \int \frac{x}{x^4+x^2+1}\ \mbox{d}x$$

$$3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x$$

## The Attempt at a Solution

I got hints but I don't know how to use them.

1) use trigonometric substitution and double angle formulas.
Do I have to use $$u=cos( \theta)$$ and after substituting the double angle formulas what to do next?

2) complete square in numerator
But I'm missing a x^2 term!

3) substitute y-1 and split integrand
Split the term?

1. I would say use x = tan(u)

plug in values for x and dx.. and simplify. Integrate with respect to u and then convert from u to x

1. I would say use x = tan(u)

plug in values for x and dx.. and simplify. Integrate with respect to u and then convert from u to x
That gives a $$\frac{1}{ \sec^4(u) }$$ does this makes life easier?

I think the 2nd one can be solved as follows:

$${\int}xdx/(x^2+1) = 1/2\sdot{\int}dx^2/(x^2+1)$$ and the rest is a piece of cake or am I missing something?

I think the 2nd one can be solved as follows:
or am I missing something?
Yes a x^4 -term.

tiny-tim
Homework Helper
1) use trigonometric substitution and double angle formulas.
Do I have to use $$u=cos( \theta)$$ and after substituting the double angle formulas what to do next?

2) complete square in numerator
But I'm missing a x^2 term!

3) substitute y-1 and split integrand
Split the term?
That gives a $$\frac{1}{ \sec^4(u) }$$ does this makes life easier?

Hi dirk_mec1!

1) Hint: $$\frac{1}{ \sec^4(u) }\,=\,cos^4u$$

2) Hint: No … you're missing a units term.

3) Do the substitution anyway, and show us what you get.

The last one:

$$3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = {\int}xde^x/(x+1)^2 = {\int} (x+1)de^x/(x+1)^2 - de^x/(x+1)^2$$

The rest I leave for you. If I am not mistaken you should get $$e^x/(x+1)$$

Sorry for my LaTeX skills. I haven't figured it out yet

Hi dirk_mec1!
2) Hint: No … you're missing a units term.

3) Do the substitution anyway, and show us what you get.

Hi tiny tim,

1) Oh now I see how to do the first one!
2) what is a units term?
3) working on it right now.

tiny-tim
Homework Helper
2) what is a units term?

Sorry … I meant an ordinary number (with no x).

Subtract a number from the denominator, to make a perfect square.

(armis, nice LaTeX, but just give hints … unless the OP didn't follow the last hint! )

yeah, sorry tiny-tim. Although often times it's kind of tricky to express my thoughts in english just verbaly but I'll try harder next time. Thanks god math notation is the same for everyone

By the way I know it's not the right place to ask but how does one use the hide feature?

Sorry … I meant an ordinary number (with no x).

Subtract a number from the denominator, to make a perfect square.
You mean like: $$(x^2+1)^2 -x^2$$

Thanks god math notation is the same for everyone
Use \frac{}{} for fractions

oh, I see. Thanks

tiny-tim
Homework Helper
You mean like: $$(x^2+1)^2 -x^2$$

No … like $$(x^2+1)^2 - 3$$ (but different, of course!)
By the way I know it's not the right place to ask but how does one use the hide feature?

hmm … I was going to ask "what hide feature?" … and then I vaguely remembered seeing something about it … I think you type the full answer, but with a tag which hides it unless someone puts the mouse over it.

But I've no idea how to operate it.

You'd better ask someone more knowledgable than me!

Anyway, it's good practice only giving hints!

No … like $$(x^2+1)^2 - 3$$ (but different, of course!)
So you mean:

$$\left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}$$

So we get

$$\int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x$$

Now...what to do next?

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tiny-tim
Homework Helper
So you mean:

$$\left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}$$

So we get

$$\int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x$$

Woohoo!
Now...what to do next?
Make the obvious substitution.

Woohoo!

Make the obvious substitution.

Let $$u = x^2+\frac{1}{2}$$

Then

$$\int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \cdot \int \frac{1}{u^2+\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \arctan \left( \frac{u}{\frac{1}{2} \sqrt{3}} \right)\ +\ C$$

Now put the u in terms of x back and we've got the answer. Is this okay Tim?

tiny-tim
Homework Helper
Yes, that's fine … except you've left out the (2/√3) outside the arctan.

(See Hootenanny's Library entry, List of Standard Integrals)

How did you get on with question 3?

I am sorry to interrupt, but wasn't there a much easier way to solve the 2nd problem?

$${\int}\frac{xdx}{(x^2+1)} = 1/2\sdot{\int}\frac{dx^2}{(x^2+1)}$$

Yes, that's fine … except you've left out the (2/√3) outside the arctan.
Whoops, you're right.

(See Hootenanny's Library entry, List of Standard Integrals)
I use the wiki-page

How did you get on with question 3?

$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = \frac{x e^x}{(x+1)^2} - \int e^x \frac{(x+1)^2-2x^2-2x}{(x+1)^4}$$

This is getting nasty....

By the way I didn't understand Armis' hint

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I am sorry to interrupt, but wasn't there a much easier way to solve the 2nd problem?

$${\int}\frac{xdx}{(x^2+1)} = 1/2\sdot{\int}\frac{dx^2}{(x^2+1)}$$

Where did you get this integral? This isn't the second integral I posted, right?

"By the way I didn't understand Armis' hint"

:) I'll explain it once you get the problem right

Ah, sorrry

I meant $${\int}\frac{xdx}{(x^2+1)^2} = 1/2\sdot{\int}\frac{dx^2}{(x^2+1)^2}$$

Oh, but it's not the same in any case :( Sorry

tiny-tim
Homework Helper
This is getting nasty....

That's because you didnt follow the nice hint:
substitute y-1

(armis … it still doesn't match the original question, does it?)

$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = \frac{-xe^x}{x+1} + \int \frac{1}{x+1} \left( e^x+x e^x \right) \ \mbox{d}x =\frac{-xe^x}{x+1} + e^x = \frac{e^x}{x+1}$$

tiny-tim
Homework Helper
$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = \frac{-xe^x}{x+1} + \int \frac{1}{x+1} \left( e^x+x e^x \right) \ \mbox{d}x =\frac{-xe^x}{x+1} + e^x = \frac{e^x}{x+1}$$

hmm … not the way they suggested … but it certainly works!

(but you'd better use their hints also, just for practice …
i] substitute y = x + 1
or ii] use partial fracitons )

hmm … not the way they suggested … but it certainly works!

(but you'd better use their hints also, just for practice …
i] substitute y = x + 1
or ii] use partial fracitons )

Ok, I found the answer with the y-1 -substitution but Tim I don't see how partial fractions is going to work, does that approach involves an (ugly) gamma function?

tiny-tim
Homework Helper
Ok, I found the answer with the y-1 -substitution but Tim I don't see how partial fractions is going to work, does that approach involves an (ugly) gamma function?

Hi dirk_mec1!

Partial fractions:

$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x$$

$$=\,\int\left( \frac{e^x}{(x+1)}\,-\,\frac{e^x}{(x+1)^2}\right) \mbox{d}x$$

which fairly obviously is:

$$\frac{e^x}{(x+1)}$$

Hi dirk_mec1!

Partial fractions:

$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x$$

$$=\,\int\left( \frac{e^x}{(x+1)}\,-\,\frac{e^x}{(x+1)^2}\right) \mbox{d}x$$

which fairly obviously is:

$$\frac{e^x}{(x+1)}$$
Thanks!

I have another integral which I can't solve (I hope you don't mind to do this in the same topic)

$$\int \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x$$

I know I have to substitute something but this thing is huge and I tried to use the seperate components of the integral for substitution but that didn't work out. Can you help me Tim?

tiny-tim
Homework Helper
Thanks!

I have another integral which I can't solve (I hope you don't mind to do this in the same topic)

$$\int \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x$$

I know I have to substitute something but this thing is huge and I tried to use the seperate components of the integral for substitution but that didn't work out. Can you help me Tim

Sorry, dirk_mec1 … x = √u doesn't seem to work, and nothing else comes to mind.

You'd better start a new thread, to get some other people's input on that one.