Question

integral x*sqrt(x^2+2x+10)

Answer 1

\[\int\limits x.\sqrt(x^2+2x+10)\]

Answer 2

please try this change of variable:
\[x ^{2}+2x+10=y\]
where y is your new variable of integration

Answer 3

ok...

Answer 4

why?

Answer 5

to make it simpler

Answer 6

oh, you mean substitution?

Answer 7

I think that the above substitution doesn't work well!

Answer 8

\[\sqrt(x^2+2x+1+9) = (x+1)^2+9\]

Answer 9

i then substitute

Answer 10

\[t = x+1 <=> t-1 = x\]

Answer 11

and dx=dt

Answer 12

that will be equivalent to integration by parts :)
if you're comfortable, then you can directly go to by parts method

Answer 13

thats where i get stuck

Answer 14

\[\int\limits_{}^{} u*dv= u*v-\int\limits_{}^{}v*du\]

Answer 15

yea, i tried

Answer 16

please, since I can write this identity,:
\[x ^{2}+2x+10=(x+2)^{2}+6\]
please try this substitution:
y=x+2
@Rowa

Answer 17

ok

Answer 18

won't that be the same thing i did?

Answer 19

oops. I have made an error,:
\[x ^{2}+2x+10=(x+1)^{2}+9\]
so y=x+1

Answer 20

yes, i've gotten to that part
the integration by parts is complicated for me

Answer 21

you should get this:
\[\int\limits (y-1)\sqrt{y ^{2}+9} dy\]

Answer 22

yes

Answer 23

now integration by parts, which is quite complicated

Answer 24

i've tried following wolframalpha, but it made things way to complicated

Answer 25

so now you can decompose that integral, namely:
\[\int\limits y \sqrt{y ^{2}+9}dy-\int\limits \sqrt{y ^{2}+9}dy\]

Answer 26

ah yes, true

Answer 27

ok! Now you can apply this substitution for the first integral:
\[y ^{2}+9=z\]

Answer 28

again? ok

Answer 29

works well for the left integral, the right one...not so much

Answer 30

you should get this:
\[\frac{ (x ^{2}+2x+10)^{3/2} }{ 3 }\]

Answer 31

as answer?

Answer 32

as final answer?

Answer 33

for first integral only!

Answer 34

oh ok

Answer 35

thats correct

Answer 36

for second integral, try this substitution, please:
\[\sqrt{y ^{2}+9}=y+z\]
where z is your new variable
please try to square both sides of that euation,

Answer 37

oops.. of that equation...

Answer 38

you should get this:
\[y=\frac{ 1 }{ 2 }\frac{ 9-z ^{2} }{ z }\]

Answer 39

ok

Answer 40

i'm sorry, but i will have to pauze this and continue later

Answer 41

so now try to differentiate that equation in order to get dy, please

Answer 42

Thank you very much, i will continue to apply your instructions when i get back

Answer 43

ok!

Answer 44

thats a pretty neat substitution \(\sqrt{x^2+bx+c} = x+t\)

Answer 45

\[\int x \sqrt{x^2+2x+10} ~dx\]
\(\sqrt{x^2+2x+10} = x+t\)

squaring we get \(2x+10 = 2xt + t^2\)

\(x = \dfrac{t^2-10}{ 2-2t}\)

the integral becomes

\[\int \frac{t^2}{8}+\frac{9}{2 (t-1)}-\frac{81}{8 (t-1)^2}+\frac{81}{4
(t-1)^3}-\frac{729}{8 (t-1)^4}+1~dt\]

which is easy to deal with

Answer 46

@Michele_Laino stuck at the second integral

Answer 47

so\[\sqrt{y ^{2}+9}=y+z\]?

Answer 48

yes!
Please square both sides, and then solve for y

Answer 49

you should get this:
\[y=\frac{ 1 }{ 2 }\frac{ 9-z ^{2} }{ z }\]

Answer 50

how did you come up with \[\sqrt{y ^{2}+9}=y+z\]

Answer 51

I write all your steps:

Answer 52

\[y ^{2}+9=(y+z)^{2}\]
so:
\[y ^{2}+9=y ^{2}+2yz+z ^{2}\]
then:
\[2yz=9-z ^{2}\]
finally:
\[y=\frac{ 1 }{ 2 }\frac{ 9-z ^{2} }{ z }\]

Answer 53

ok thanks, but why \[\sqrt{y ^{2}+9}=y+z\]

Answer 54

i get the first part, the y+z is difficult to understand

Answer 55

it is technique of integration!

Answer 56

oh ok, i will have to learn it

Answer 57

that's right!

Answer 58

now, please differentioate that formula above, in order to get dy, namely:
\[dy=...dz\]

Answer 59

you should get this:
\[dy=-\frac{ 1 }{ 2 }\frac{ z ^{2}+9 }{ z ^{2} }dz\]

Answer 60

thats quick, ok

Answer 61

ok! now, please substitute the expressions which we have found, namely y, dy and
sqrt(y^2+9) as function of z, into your integral

Answer 62

you should get this:
\[-\frac{ 1 }{ 4 }\int\limits \frac{ (9+z ^{2})^{2} }{ z ^{3} }dz\]

Answer 63

what is sqrt((y^2+9) equal to, i mean substituted as

Answer 64

oops...note that:
\[\sqrt{y ^{2}+9}=y+z=\frac{ 9-z ^{2} }{ 2z }+z=\frac{ 9+z ^{2} }{ 2z }\]

Answer 65

i think i'm lost

Answer 66

where?

Answer 67

when you used sqrt(t^2+9) = t+z

Answer 68

i thought you were substituting at first..

Answer 69

it is the same, I do that substitution now

Answer 70

then what is y equal to, if you want to substitute it