# More Integration Problems

• duki

## Homework Statement

Sorry for all the posts; I'm trying to study for a test tomorrow.

1. Integrate
$$\int\frac{dx}{x^2+8x+65}$$

2. Integrate
$$\int\frac{2x^3}{\sqrt(x^2+4)}dx$$

## The Attempt at a Solution

1.
$$u = x^2 + 8x + 65$$
$$du = 2x+8dx$$
Stuck.

2.
$$\int2x^3(x^2+4)^{-\frac{1}{2}} dx$$
$$\frac{2x^4}{4}\frac{x^2+4}{\frac{1}{2}} + C$$
Not sure if this is right so far.

Thanks for any help!

For the first one, a simple u-substitution won't work. Why not try completing the square?

For the second one, think about trig substitution.

Integrals that need either of these tricks tend to have pretty recognizable forms, so you should familiarize yourself with which forms need which trick.

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Also, with #1:

I know it has something to do with the rule $$\int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan \frac{u}{a} + C$$

For #1 I think you want to write x^2 + 8x + 65 in the form a + (x+b)^2, then you'll have $$\int \frac{1}{a+(x+b)^2} = \frac{1}{a} \int \frac{1}{1 + (x+b)^2/a}$$ which you can then use the substitution u^2 = (x+b)^2/a so that you'll end up with an inverse tangent.

#1 - Complete the square

$$\int\frac{dx}{x^2+8x+16+49}$$
$$\int\frac{dx}{(x+4)^2+49}$$ ?

So then

$$u = x+4$$
$$du = dx$$
$$\int\frac{du}{u^2+7^2} = \frac{1}{7}\arctan\frac{u}{7} + C = \frac{1}{7}\arctan\frac{x+4}{7} + C$$
?

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duki said:

$$\int\frac{dx}{x^2+8x+16+49}$$
$$\int\frac{dx}{(x+4)^2+49}$$ ?

So then

$$u = x+4$$
$$du = dx$$
$$\int\frac{du}{u^2+7^2} = \frac{1}{7}\arctan\frac{u}{7} + C = \frac{1}{7}\arctan\frac{x+4}{7} + C$$
?

Looks good

awesommeeee
I'm still stuck on the 2nd one tho..

What sort of trig substitution?

#2 - Trig sub.

$$x=2\tan\theta$$

duki said:
What sort of trig substitution?

There's a standard form each trig substitution. The expression under the radical is of the form $$x^{2} + a^{2}$$. Your book should tell you which trig function to substitute.

CrazyIvan said:
There's a standard form each trig substitution. The expression under the radical is of the form $$x^{2} + a^{2}$$. Your book should tell you which trig function to substitute.

Too late, Roco did. But you don't really need the book to tell you. What sort a trig function can you square and add a constant and get something you can extract the square root of?

Dick said:
Too late, Roco did. But you don't really need the book to tell you. What sort a trig function can you square and add a constant and get something you can extract the square root of?

No clue. =(
I found $$\int\frac{du}{\sqrt{a^2-u^2}} = \arcsin \frac{u}{a} + C$$
Is this what I need?

That has a negative sign.

I don't know. I'm fixing to go to bed to get a good nights sleep though, is there any way you could give me another clue to what I should be looking for?

duki said:
No clue. =(
I found $$\int\frac{du}{\sqrt{a^2-u^2}} = \arcsin \frac{u}{a} + C$$
Is this what I need?

Sorry, I'm distracting you. tan(x)^2+1=sec(x)^2. Rocophysics already told you the correct substitution.

haha, I thought someone might say that... I don't know how to use his substitution.
The answer I was given to check myself is $$\frac{2}{3}(x^2+4)^{\frac{1}{2}}(x^2-8) + C$$

That's it alright. Sleep on it and check it tomorrow.

lol
I can't. My test is tomorrow D=
Not sure if this type is going to be on there, but I was hoping to get it down just in case.

Let me try one more time. Can you tell me where I'm going wrong? From the beginning?

$$\int\frac{2x^3}{\sqrt{x^2+4}}dx$$
$$u = 2x^3$$
$$du = 6x dx$$
$$\int\frac{u}{\sqrt{x^2+4}}\frac{du}{6x}$$

right so far?

Ok, you don't need a trig substitution for this one. Try u=x^2+4. That leaves you with an extra x^2 in the numerator. But that's ok. x^2=u-4.