Question

Calculus:
I have a test in 4 hours and I need to learn this stuff quick. If you could lay out the steps and a breif examplantion so I can follow your rationale that would be great.

3. Find the following indefinite integrals. If you use integration by substitution, write what you
chose to be u and find du. Don't forget to add the constant!

(a) {integral sign} [x-root(3x2 + 1)] dx

(b) {integral sign} [(4+lnx)^2/x] dx

(c) {integral sign} [(x^2e^(6-x^3)] dx

(d) {integral sign} [(2x^3-3x)/(x^4-3x^2+1)] dx

Answer 1

@lgbasallote

Answer 2

for number 1 is that \[\int x\sqrt{3x^2 + 1}dx\]

Answer 3

@MercerC you here?

Answer 4

Yes it is

Answer 5

okay...here's a hint..try letting \[u = 3x^2 + 1\]
what do you get as du then?

Answer 6

nevermind I think I'm just going to repeat this class. this is overwhelming

Answer 7

i was just doing integration by substitution

Answer 8

to find du you just take the derivative of \(3x^2 + 1\)

Answer 9

6x

Answer 10

right. \[du = 6xdx\]
so \[xdx = \frac{du}{6}\]
agree? i just divided both sides by 6

Answer 11

okay

Answer 12

so now..our integral was \[\int x\sqrt{3x^2 + 1} dx\]
but i can also rearrange it like \[\int (\sqrt{3x^2 + 1})xdx\]
right? commutative property of multiplication?

Answer 13

okay

Answer 14

so...remember i said \[u = 3x^2 + 1\]
and \[\frac{du}{6} = xdx\]
so i can change the integral to \[\int (\sqrt{3x^2 + 1} )xdx \implies \int (\sqrt u) \frac{du}{6}\]
following so far?

Answer 15

yep

Answer 16

now..i'll take 1/6 out of the integral \[\frac 16 \int (\sqrt u) du\]
then change \(\sqrt u\) into \(u^{1/2}\)
\[\implies \frac 16 \int (u^{1/2})du\]
okay?

Answer 17

following...

Answer 18

good...now do you remember the rule in integral that looks like this.. \[\Large \int x^n \text dx \implies \frac{x^{n+1}}{n+1}\]
does that seem familiar?

Answer 19

nope

Answer 20

I'm sorry, I'm going to stop you. Thank you for your help. But I think there is just way too much of this test that I don't know.

Answer 21

this is called the power formula..it is used a lot when integrating...basically when you have an integral that looks like \[\int x^n dx\] the integral of that is just \[\frac{x^{n+1}}{n+1}\]
x is the variable and n is the exponent.

in this problem we have
\[\int u^{1/2} du\]
so the integral of that is \[\large \frac{u^{\frac 12 + 1}}{\frac 12 + 1}\]
make sense?

Answer 22

an accellerated calculus class was a bad idea

Answer 23

do you have background knowledge in calculus?

Answer 24

nope, its the only course I need to take to graduate.why is my solution this:

Answer 25

\[1/9(3x ^{2}+1)^{3/2}+C\]

Answer 26

well like i was saying \[\Large \frac{u^{\frac 12 + 1}}{\frac 12 + 1} \implies \frac{u^{\frac 32}}{\frac 32} \implies \frac{2u^{\frac 32}}{3}\]
then you had a 1/6 outside so \[\Large \frac 16 \times \frac{2u^{\frac 32}}{3} \implies \frac 13 \times \frac{u^{\frac 32}}{3} \implies \frac{u^{\frac 32}}{9}\]
then you substitute u back into \(u = 3x^2 + 1\)
\[\Large \frac{u^{\frac 32}}{9} \implies \frac{(3x^2 + 1)^{\frac 32}}{9}\]
sorry if it looks complicated :(

Answer 27

It's okay. I'm sure its easy for you. I'm just having a tough time understanding what the heck I'm learning. It's all so abstract and not relevant to any of my education. I wish I would have taken this 4 years ago when suggested.

Answer 28

believe it or not...it took me two tries to pass the course...that's why i know it well now..

Answer 29

Me too, this accelerated course didn't really teach me much. It was just an overflow of information and terminology that I don't understand.

Answer 30

right now, I'll probably get a 15% on the final. So, I'm thinking of calling it a night and trying next semester.

Answer 31

thanks for your help.

Answer 32

actually...i passed my second try in calculus because of this site :D this site taught me a lot in calculus back in the days