Question

integral of x^2lnx

Answer 1

i've just done it now and got\[\frac{ \frac{ x ^{3}}{3} lnx-\frac{ 1 }{ 3 } (x^{3}lnx)}{4 }\]

Answer 2

\[x^2 \cdot \ln(x)\]

Answer 3

you can use Integration By Parts here.. :)

Answer 4

you need to do integration by parts (twice)

Answer 5

ive done by parts twice

Answer 6

Wait, let me reach where you have reached.. :P

Answer 7

my options are do by parts twice and get to the above answer or simplify the x^3/x to x^2 and integrate 1/3Sx^2

Answer 8

well, I will take \(\ln(x)\) as first function and \(x^2\) as second..

Answer 9

your second opinion is looking good to me.

Answer 10

the one where i simplify the x?

Answer 11

\[\int\limits x^2 \cdot \ln(x) \cdot dx = \ln(x) \cdot \frac{1}{3}x^3 - \int\limits \frac{1}{x} \cdot \frac{1}{3}x^3 \cdot dx\]

Answer 12

There is no need to use one thing twice.. :P

Answer 13

\(x\) and \(x\) cancels there and you are left with \(x^2dx\) in the second integral after taking 1/3 outside of integral.

Answer 14

i see yes. but if i were to do it accidentally would it be wrong?

Answer 15

Actually, can you tell what is integral of \(\ln(x)\) ?/

Answer 16

i can't but it will be on my appendix

Answer 17

Then if you don't know or don't tell, then how can you do two times by parts there?

Answer 18

Because if you take \(\ln(x)\) as your first function, then you must know its integral too.

Answer 19

We know the derivative of \(\ln(x)\) easily, but its integration, I also don't know this time.. :P

Answer 20

*\(\ln(x)\) as second function..

Answer 21

i let u=lnx du=1/x
and dv=x^2 v=x^3/3

Answer 22

If you are taking \(u\) as \(ln(x)\), then you have to use parts only one time.

Answer 23

Have you read the word "ILATE" anywhere while studying?

Answer 24

ive heard LIATE from my lecturer

Answer 25

LIATE or ILATE?

Answer 26

LIATE

Answer 27

LIATE = Log Inversetrig Algebraic Trig Exponential

Answer 28

Perl.

Answer 29

In India, Inverse is given more preference, may be the country name is also starting from I too.. :P

Answer 30

ILATE : Inverse Trig, Log, Algebraic, Trig, Exponential.

Answer 31

sorry to jack my own question buyt is this other answer correct

Answer 32

haha i think so too

Answer 33

Okay, in both cases, Log has preference over Algebraic, so ILATE or LIATE is basically a convention, it is not a rule, remember this, while solving by parts, use this convention but it is not at all necessary, you can take any function as u and v, but this convention simplifies some of the integrals, like this one..

Answer 34

Sorry, what answer you are talking of?

Answer 35

the one in the attached picture

Answer 36

How does \(e^{3x}\) become \(e^x\) ??

Answer 37

i forgot to write the 3 x there but i fixed it in the next line

Answer 38

NO..

Answer 39

In the last line, you wrote \(e^x\) there.

Answer 40

i see yes, i meant 3x

Answer 41

See, I appreciate, that you know how to use the rule or formula, but you are doing silly mistakes while simplifying the integral.

Answer 42

Okay, quickly let us do once again, you are writing up to:

\[x^2 \cdot \frac{1}{3}e^{3x} - \frac{2}{3} \int\limits e^{3x} x \cdot dx\]

I have taken 2 from (2x) outside too. Okay??

Answer 43

*you are correct up to.... :)

Answer 44

Okay??

Answer 45

yes

Answer 46

Forgot all the term up to 2/3 there, just integrate it at separate place:

\[\int\limits e^{3x} \cdot x \cdot dx\]

Answer 47

*Forget

Answer 48

ill do that quickly

Answer 49

Don't go with the speed of light, be slow but accurate.. :)

Answer 50

Good.. :)

Answer 51

Now, we have to fit this there in our actual problem, but be careful.. I do it for you :

Answer 52

\[x^2 \cdot \frac{1}{3}e^{3x} - \frac{2}{3} \int\limits\limits e^{3x} x \cdot dx = x^2 \cdot \frac{1}{3}e^{3x} - \frac{2}{3}[\frac{x \cdot e^{3x}}{3} - \frac{e^{3x}}{9}] + C\]

Correct?

Answer 53

yes i got that one now

Answer 54

Clear now??

Answer 55

yes thank you

Answer 56

you are welcome dear.. :)