Question

lntegral attack! will fan and medal and build you a crown!

Answer 1

\[\int\limits xe^(x^2+2)^1/3\]

Answer 2

it's xe^(x^2+2)^1/3

Answer 3

I did write x^2+2=u

Answer 4

and it's really messy from there

Answer 5

I get lost everytime

Answer 6

@rvc?

Answer 7

@Andras ?

Answer 8

I will write down what I've come through so far

Answer 9

\[x^2+2=U, du=2x\]

Answer 10

xdx=du/2

Answer 11

\[0.5\int\limits e^u^1/3\]

Answer 12

does it make sense so far?

Answer 13

@rvc?

Answer 14

@IrishBoy123 ?

Answer 15

@IrishBoy123 PELASE

Answer 16

that look sabsolutely horrible.

have you tried u = (x^2 + 2) ^ (1/3)? i mean really rip the guts out of it.....

i'll have a go that way.

Answer 17

it works!!!!!

it needs by parts a coupla times as you get a squate term but it looks good.

u^3 = x^2 + 2
3 u^2 du = 2 x dx

Answer 18

I tried so hard! I had a monnlit night because of that!

Answer 19

I gtg to work, I'm begging please help!

Answer 20

following on

dx = (3 u^2 / 2x) du

that gives putting it all together: x (e^ u) (3) (u^2) / 2x du

== ∫ 3 u^2 e^u du


yes?!?!

typing on hoof here but this shld be good

Answer 21

@IrishBoy123 the answer is \[\frac{ 3 }{ 2 }e^\sqrt[3]{x^2+1}(\sqrt[3]{(x^2+1)^2}-2\sqrt[3]{x^2+1}+2)+C\]

Answer 22

Sorry it took so long, I've just come back from work

Answer 23

@rvc ?

Answer 24

@IrishBoy123 I'm afraid it's not the way...

Answer 25

i havent actually done it, just set it up.

hope its right, well done!

Answer 26

??? you found another way??

Answer 27

not yet, it took my hours of sleep away, by hook or by crook, I shall solve it today!

Answer 28

i'll have time later to finish my suggestion....good luck in meantime!

Answer 29

thank you so much! I'll keep you updated, thanks for everything btw!

Answer 30

@SithsAndGiggles HELP!

Answer 31

@perl

Answer 32

$$
\Large \int\limits xe^{{(x^2+2)} ^{1/3}}$$

Answer 33

is that the question?

Answer 34

yes!

Answer 35

u = x^2 + 2
du = 2x dx

du/2 = x dx

ok so far?

Answer 36

yes!

Answer 37

$$
\Large{
\int\limits xe^{{(x^2+2)} ^{1/3}}~dx\\
\\ \iff \\
\int\limits e^{{(x^2+2)} ^{1/3}}x~dx\\
\\ ~\\
u = (x^2 + 2)\\
du = 2x dx\\
\frac{du}{2} = x dx

\\ \iff \\
\Large \int\limits e^{{u} ^{1/3}} \frac{du}{2}


}

$$

Answer 38

so far so good

Answer 39

this function has no elementary antiderivative

Answer 40

oh no!

Answer 41

the antiderivative cannot be expressed as an elementary function

Answer 42

then what shall we do?!

Answer 43

what are the directions again, dont leave out anything

Answer 44

hmmm one sec

Answer 45

by solved i mean find an antiderivative in terms of elementary functions ( logs, inverse trig, algebraic, trig, exponential)

Answer 46

http://www.symbolab.com/solver/step-by-step/%5Cint%20xe%5E%7B%5Cleft(x%5E2%2B2%5Cright)%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D/?origin=button

Answer 47

That's according to symbolab

Answer 48

so it is somehow possible, but idk why...

Answer 49

ok lets do the substitution
u^3 = x^2 + 2

Answer 50

u^3?

Answer 51

you mean u^(1/3)?

Answer 52

u^3 = x^2 + 2
3u^2 du = 2x dx

do you see why this is useful substitution.
now
u = (x^2 + 2)^(1/3)

Answer 53

but how did you even come up with this idea?

Answer 54

you want the integral to contain e^(u), since you know the antiderivative of e^u.
so by cubing the power of u when you do the u substitution, you anticipate the cube root

Answer 55

ok! that's a start!

Answer 56

by the way, you don't have to do it this way, but it seems like the simplest. i will show you how the program you linked is doing it, which is pretty cool as well

Answer 57

lets call the substitution u^3 = x^2 + 2 , the irish substitution, since he suggested above.
or we can do the symbolab approach

Answer 58

i'll latex both ways, ok ?

Answer 59

thanks a lot!

Answer 60

I really appreciate that!

Answer 61

does that make sense so far?

Answer 62

now you can do integration by parts , twice

Answer 63

I didn't get how the substitute v2=(u1/3)2

Answer 64

why to do v^2?

Answer 65

do you understand what I'm asking?

Answer 66

the problem is we have that extra 3u^(2/3) , and we only want an integral in v and dv

Answer 67

so we go back to the substitution
v = u^1/3

now if we square both sides, we have an expression of u^2/3 in terms of v

Answer 68

wow

Answer 69

that's just insane

Answer 70

I hope I'll take it from here. Thanks a lot!

Answer 71

$$
\Large{
\int\limits xe^{{(x^2+2)} ^{1/3}}~dx = \int\limits e^{{(x^2+2)} ^{1/3}}x~dx\\
\\ ~\\
u = (x^2 + 2), ~~ du = 2x~ dx, ~~\frac{du}{2} = x dx

\\ \therefore \\
\int\limits e^{{(x^2+2)} ^{1/3}}x~dx= \Large \int\limits e^{{u} ^{1/3}} \frac{du}{2}=
\frac 12\Large \int\limits e^{{u} ^{1/3}}~ du
\\ ~ \\
v = u^{\frac13},~~ dv = \frac {1}{3u^{2/3}} ~du, ~~~3u^{2/3}~dv = du
\\\\\frac 12\Large \int\limits e^{{u} ^{1/3}}~ du
= \frac 12 \int e^v~ \color{red}{3u^{2/3}}~dv

\\ \text{But...} ~~ \color{blue}{u^{2/3} =(u ^{1/3})^2 = (v)^2}

\\ \therefore \\
\\ \frac 12\Large \int\limits e^{{u} ^{1/3}}~ du = \frac 12\Large \int\limits e^{v}~ 3v^2 dv
= \frac 3 2 \int v^2 e^v dv

}

$$

Answer 72

i wanted to show you irish's substitution suggestion, less steps

Answer 73

really?

Answer 74

yes its more elegant

Answer 75

I would love that! PLEASE!

Answer 76

$$
\Large{
\int\limits xe^{{(x^2+2)} ^{1/3}}~dx= \int\limits e^{{(x^2+2)} ^{1/3}}x~dx\\
\\ ~\\
u^3 = (x^2 + 2), ~~ 3u^2 du= 2x~dx, ~\frac 32 u^2 du= x~dx \\

\\ \therefore \\
\Large \int\limits e^{{(x^2+2)} ^{1/3}}x~dx = \int e^{(u^{1/3})^3}~\frac 32 u^2 du

\\ = \frac 32 \int u^2 e^u du
}
$$

Answer 77

thats the same integral we got above , but with an extra substitution steps, so we ended up with an integral in v

Answer 78

step*

Answer 79

@perl you're truly sagacious!

Answer 80

thanks, nice link by the way :)

Answer 81

you don't understand, I've been working on it for too long!