Question

anyone help me with integration by parts

Answer 1

i can try

Answer 2

just a moment. let me post the question pl.

Answer 3

integrate Z^n ln z. dz

Answer 4

@im.celibate

Answer 5

use LIATE method
so
logarithm=first function
algebric=second function
\[\int\limits_{}^{}z^n.\ln z~dz=\ln z.\int\limits_{}^{}z^ndz-\int\limits_{}^{}[\frac{ d }{ dx }(\ln z)*\int\limits_{}^{}z^ndz]dz\\=\ln z.\frac{ z^{n+1} }{ n+1 }-\int\limits_{}^{}\frac{ 1 }{ z }*\frac{ z^{n+1} }{ n+1 }dz\\=\ln z.\frac{ z^{n+1} }{ n+1 }-\frac{ 1 }{ n+1 }*\int\limits_{}^{}z^n dz\]
i believe you can do the rest now!!! :)
@ankitben

Answer 6

Thanks a lot.

Answer 7

i have another question

Answer 8

divide the polynomial X^2-1/ X-1

Answer 9

x^2-1=(x+1)(x-1)

Answer 10

ok?

Answer 11

Thank you. one more question,

Answer 12

please hang on

Answer 13

Show that if a and b are relatively prime numbers (the greatest common divisor is 1),
and ab is a square, then both a and b are squares. [Hint: use the Fundamental Theorem
of Arithmetic].

Answer 14

last qns please

Answer 15

and X^5- 1 / X-1

Answer 16

\[x^5-1=(1+x+x^2+x^3+x^4)(x-1)\]
use this

Answer 17

thank yo and for the statement problem?

Answer 18

yes check this out
http://math.stackexchange.com/questions/284636/if-a-and-b-are-relatively-prime-and-ab-is-a-square-then-a-and-b-are-squares

Answer 19

awesome. Thanks Sid. This one too pl. Show that if a and b are relatively prime numbers (the greatest common divisor is 1),
and ab is a square, then both a and b are squares. [Hint: use the Fundamental Theorem
of Arithmetic].

Answer 20

sorry not this

Answer 21

Note that 19 = 4 5 􀀀 1; 23 = 4 6 􀀀 1. Prove that there are innitely many primes of
the form 4k 􀀀 1 where k 2 Z

Answer 22

question isnt clear

Answer 23

let me type it.

Answer 24

Note that 19= 4.5-1 and 23= 4.6-1 prove that there are infinitely many primes of the form

Answer 25

4k-1 where k belongs to Z

Answer 26

Sid?

Answer 27

yeah wait lost connection

Answer 28

Thank you.

Answer 29

really dont understand the question :(

Answer 30

Its okay, Thank you. i have another qns [X^n+1]-1 / X-1

Answer 31

n + 1 goes in the power

Answer 32

wait its lagging very much and i'm loosing connection again and again

Answer 33

yea sure. Take ur time.

Answer 34

\[\frac{ x^{n+1}-1 }{ x-1 }=\sum_{i=0}^{i=n}x^n=1+x+x^2+x^3+x^4+.....+x^n\]

Answer 35

Thats it ?

Answer 36

yes thats the standard series
\[x^{n+1}-1=(x-1)(1+x+x^2+x^3+....+x^n)\\so \\ \frac{ x^{n+1}-1 }{ x-1 }=1+x+x^2+....+x^n\]

Answer 37

than you so much Sid.