Question

make a substitution to express the integrand as a rational function and then evaluate the integral

how do i do this? problem inside!! thanks!!

Answer 1

\[\int\limits_{} \frac{ dx }{ 2\sqrt{x+3}+x }\]

Answer 2

What have you tried?

Answer 3

i am unsure of which concept i am using and how to apply

am i doing the u= ____ du=____ ?

Answer 4

Yes, when they ask for a substitution you can make u = something

Answer 5

okay, would it be u=2 (sqrt x+3) +x and du = dx ?

Answer 6

i am very confused :(

Answer 7

Yes that's a good substitution! But when you have to take the derivative with respect to x that's why you have \[\frac{ du }{ dx }\]

Answer 8

ohh okay, so what would i be using as substitution in this case then?

Answer 9

can i use this?

u=2 (sqrt x+3) +x and du = dx

?

Answer 10

You will be using \[u=\sqrt{x+3}\] what's the derivative of this?

Answer 11

would it be

1
-----
2 sqrt(x+3)

?

Answer 12

and if so, that would be du?

Answer 13

Yup, but note we have that +x in the denominator so you have solve for x here as well \[u=\sqrt{x+3}\] solve for x

Answer 14

\[\sqrt{x+3}=u,x+3=u^2,x=u^2-3,dx=2udu\]

Answer 15

\[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]

Answer 16

ooh okie!! so what happens next? :/

Answer 17

do i still need to find more substitutions?

Answer 18

Well I asked you to solve for x, but surji already did it for you... now just plug it all in.

Answer 19

ohh okay, so i plug into the original equation?

Answer 20

Yes, that's why it's called substitution :)

Answer 21

like this? \[\int\limits_{} \frac{ 2udu }{ 2\sqrt{(u^2-3)+3}+(u^2-3) }\]

Answer 22

Nope, lets go over this slowly

Answer 23

aww:( okay!!

Answer 24

So our original integral is \[\int\limits \frac{ dx }{ 2\sqrt{x+3}+x }\] we made the substitution \[u= \sqrt{x+3} \implies du = \frac{ 1 }{ 2\sqrt{x+3} }dx\] but notice the integrand still has that +x in the denominator, and we can't have that when we want everything respect to u, so we look at our original substitution and see if there is a way we can get rid of the x, \[u=\sqrt{x+3} \implies x = u^2-3\] so we plug that in where that +x is, but note we can simplify our derivative \[du = \frac{ 1 }{ 2\sqrt{x+3} }dx \implies du = \frac{ 1 }{ 2u }dx \implies 2udu = dx\]

Answer 25

Now try and plug it all in

Answer 26

whoahh okay, let me try :)

Answer 27

\[\int\limits_{} \frac{ 2udu }{ 2u+(u^2-3) }\]

Answer 28

this?

Answer 29

Yes, exactly!

Answer 30

ooh okay yay!! so what do i do now?

Answer 31

Now you can evaluate it

Answer 32

how can i do that? :/ sorry, i am having a hard time understanding calc :( i follow so far right now, but i do not understand what I am supposed to do when you say evaluate? is it finding the antiderivative?

Answer 33

Yes, when you integrate you find the antiderivative.
You will have to use partial fractions

Answer 34

okay, i am not very good with partial fractions :( could you possibly guide me through the steps that I would do here? I just don't know what i do here :/ find A,B and C?

Answer 35

Well partial fractions can only be learnt through practice, you have to practice in calculus if you want to pass, so you should practice these problems after.

Lets start of by looking at the integrand \[2 \int\limits \frac{ u }{ u^2+2u-3 }du\] lets factor the denominator what do you get?

Answer 36

i get (u+1)-1 ? :/

Answer 37

Factor the denominator u^2+2u-3

Answer 38

okay, so (u+1)(u-4) ? :/

Answer 39

(u+1)^2 + 1 ?

Answer 40

i don't really know :(

Answer 41

Ok so you will have to review factoring as well, you do this in high school, you should have got (u+3)(u-1)

Answer 42

So we have now \[2 \int\limits \frac{ u }{ (u+3)(u-1) }du\] this is our integral now

Answer 43

So for partial fractions we set it up as such \[\frac{ u }{ (u+3)(u-1) } = \frac{ A}{ (u+3) }+\frac{ B }{ (u-1) }\]

Answer 44

ohhh okay, oops, sorry thought i had to do the thing where you add a 1 to the end or something :/

Answer 45

okay! and how do we solve for A and B ?

Answer 46

No you're thinking of completing the square, which is also helpful for some problems

Answer 47

ohh okay, oopsies :P

Answer 48

Multiply both sides by (u+3)(u-1) what do you get

Answer 49

A(u-1) and B(u+3) ?

Answer 50

Yes, but you must write it as \[u=A(u-1)+B(u+3)\]

Answer 51

Now distribute the A and B

Answer 52

ohh okay!! distribute like Au-A + Bu+3B ?

Answer 53

Yeah, now we look at what we have on the left side, \[u=(A+B)+(-A+3B)\] now we can set up a system of equations \[A+B=1\]
\[-A+3B=0\] I think you should be able to do the rest

Answer 54

okay!! err, still a bit confused :/ what would i be doing next? do i sub in to the A/u+3 + B/u-1 ?

Answer 55

with A=1 and B=0? :/

Answer 56

This is just a regular system of equations solve for A and B

Answer 57

You can use any method, I recommend just adding as it's the most obvious in this case

Answer 58

ok, so we get A=3B --> 3B+B =1 =4B = 1 -->B=1/4

then A=3/4 ?

Answer 59

did i do that correctly?

Answer 60

Yeah, looks good!

Answer 61

yay! what do i do now?

Answer 62

\[\frac{ u }{ (u+3)(u-1) } = \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }\] now this will become your integral so we have \[2 \int\limits \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }du\] this should be easy now, just evaluate

Answer 63

so would i get this?\[\frac{ 1 }{ 2 } \log(1-u) + 3\log(u+3) + C \]

Answer 64

\[2(3/4\ln|u+3|+14\ln|u-1|)+C\] but you have to sub back the u substitution we made

Answer 65

1/4 not 14

Answer 66

okay! so we get this? \[2(3/4\ln \left| 5/4+3 \right|+ 1/4\ln \left| 5/4-1 \right| +C\]

Answer 67

5/4?

Answer 68

Curious, where did you get the 5/4 from?

Answer 69

i am not quite sure if i calculated correctly?

u=A+B -A +3B ?

=(3/4)+(1/4)-(3/4)+3(1/4)
=1 ? oops :P

Answer 70

Noooo, all we have to do is go back and see \[u= \sqrt{x+3}\] \[\implies 2\left( \frac{ 3 }{ 4 }\ln|\sqrt{x+3}+3|+\frac{ 1 }{ 4 }\ln|\sqrt{x+3}-1| \right)+C\]

Answer 71

ohhh i see

so what do we do next?

Answer 72

We're done, the problem not long enough? :P

Answer 73

hahaha ohhh okay :P yay!! thank you so ugh!! it was so long!! but i think i understand more of it now :) thanks so much!!! I'm having trouble understanding the concepts for calc right now so i appreciate your time and help!!

Answer 74

Np, integrals require lots of practice, I haven't mastered them myself xD, they can be tricky, they're not as easy as derivatives!
So keep practicing!

Answer 75

:)