1. Originally Posted by chocolatelover

For the int. 2/x I got 2ln(x), for the int. x/x sq.+3 I got ln(x) and I'm not sure what to do for the int. 1/x sq. +3

Can you give me a hint or show me what to do, please?

Thank you
your answer for $\displaystyle \int \frac {x}{x^2 + 3}~dx$ is wrong. do it again. use substitution, $\displaystyle u = x^2 + 3$

for $\displaystyle \int \frac {1}{x^2 + 3}~dx$ we proceed as follows.

note that: $\displaystyle \frac {1}{x^2 + 3} = \frac {1}{3 \left( \frac {x^2}{3} + 1 \right)}$

$\displaystyle = \frac {1}{3} \cdot \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }$

so, $\displaystyle \int \frac {1}{x^2 + 3}~dx = \frac {1}{3} \int \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }~dx$

now continue

2. Thank you

This is a u substitution too, right? Then you would do integration by parts, right?

For the second interval I got 1/2ln|u|. Is that correct?

Thank you

3. Originally Posted by Jhevon
$\displaystyle \int \frac {1}{x^2 + 3}~dx = \frac {1}{3} \int \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }~dx$
Wouldn't be faster if we take

$\displaystyle \int\frac1{x^2+3}\,dx=\int\frac1{x^2+\left(\sqrt3\ right)^2}=\frac1{\sqrt3}\arctan\frac x{\sqrt3}+k?$

4. Originally Posted by Krizalid
Wouldn't be faster if we take

$\displaystyle \int\frac1{x^2+3}\,dx=\int\frac1{x^2+\left(\sqrt3\ right)^2}=\frac1{\sqrt3}\arctan\frac x{\sqrt3}+k?$

yes, but that's basically memorizing a formula. my memory is horrible, so i always work things out the long way

5. Originally Posted by Jhevon
yes, but that's basically memorizing a formula.
Not hard to memorize

Actually, this helps when you're giving a test at the college, isn't?

6. This is a u substitution too, right? Then you would do integration by parts, right?

For the second interval I got 1/2ln|u|. Is that correct?

7. Originally Posted by chocolatelover
This is a u substitution too, right? Then you would do integration by parts, right?

For the second interval I got 1/2ln|u|. Is that correct?
what integral are you talking about?

there is no by parts here. regular substitution or a memorized formula will get you through

Originally Posted by Krizalid
Not hard to memorize
that one is not hard to memorize, but it is one of many. if i memorize that one, i might as well memorize the one for arcsine, arccosine ...where does it end?!

Actually, this helps when you're giving a test at the college, isn't?
well, this level of algebraic manipulation doesn't slow me down really. at least not enough to warrant memorizing formulas that it is not the end of the world if i don't memorize them

8. For this one do I use integration of parts? Is the integral x/x sq. +3 1/2ln|u|?

9. Originally Posted by chocolatelover

For this one do I use integration of parts? Is the integral x/x sq. +3 1/2ln|u|?
it is not integration by parts. you use the substitution $\displaystyle u = \frac {x}{\sqrt {3}}$ and you will get the answer that Krizalid got by using the formula. see the previous page in this thread

the answer is $\displaystyle \frac1{\sqrt3}\arctan\frac x{\sqrt3}+C$

10. Is du just dx then?

So, it would 1/3int. udu=

1/3int.u=

1/3 u2/2

that means that=

1/3(x/sq. 3)2

How does that equal 1/3arctanx/sq.3?

11. Try to get it more generally

$\displaystyle \int\frac1{a^2+u^2}\,du$

Apply trig. substitution defined by $\displaystyle u=a\tan\theta$

12. Can you show me what you mean? My book shows me how to do this type of problem, but it skips a lot of steps and I just started learning this.

Thank you

13. Originally Posted by chocolatelover
Can you show me what you mean? My book shows me how to do this type of problem, but it skips a lot of steps and I just started learning this.

Thank you
$\displaystyle \int \frac{1}{a^2 + u^2} ~ du$

Let $\displaystyle u = a \cdot tan(\theta) \implies du = a \cdot sec^2(\theta) d \theta$

Thus
$\displaystyle \int \frac{1}{a^2 + u^2} ~ du = \int \frac{1}{a^2 + a^2tan^2(\theta)} \cdot (a \cdot sec^2(\theta)) d\theta$

$\displaystyle = \int \frac{a}{a^2}\frac{sec^2(\theta)}{1 + tan^2(\theta)}~ d \theta$

Now, $\displaystyle 1 + tan^2(\theta) = sec^2(\theta)$, which is the whole motivation for the form of the substitution, so
$\displaystyle = \frac{1}{a} \int \frac{sec^2(\theta)}{sec^2(\theta)}~d \theta$

$\displaystyle = \frac{1}{a} \int ~d \theta$

$\displaystyle = \frac{1}{a} \cdot \theta + C$

And $\displaystyle u = a \cdot tan(\theta) \implies \theta = tan^{-1} \left ( \frac{u}{a} \right )$, so finally:

$\displaystyle \int \frac{1}{a^2 + u^2} ~ du = \frac{1}{a} \cdot tan^{-1} \left ( \frac{u}{a} \right ) + C$

-Dan

14. Originally Posted by topsquark
$\displaystyle = \frac{1}{a} \cdot \theta + C$

And $\displaystyle u = tan(\theta) \implies \theta = tan^{-1}(u)$, so finally:

$\displaystyle \int \frac{1}{a^2 + u^2} ~ du = \frac{1}{a} \cdot tan^{-1}(u) + C$
You missed something in the second line

15. Originally Posted by Krizalid
You missed something in the second line
Two errors this morning and at least one last night! I'm off my game. Thanks for the spot. I fixed it in the original post.

-Dan

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