how to integrate logarithmic function? for eg:- log(1+x)dx from 0 to 1 ?

Question

it says i will get answer log(4/e)

i dont know how do i get it

using a highpower scientific calculator
(ms 570 casio) i get it numerically correct but still i dont know how to solve it

Answer 1

The previous person almost got there.
The indefinite integral is (1+x)[log(1+x) - 1] + C

Substitute x=1 to get 2[log(2) - 1] + C
Substitute x=0 to get -1 + C.

The definite integral is
2[log(2) - 1] - [-1]
= 2log(2) -1
=log(4) - 1
=log(4) - log(e)
=log(4/e)

Answer 2

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Answer 3

by parts intergration take(1+x)=t
then dx=dt

int logtdt=tlogt-int t*1/tdt
=tlogt-t+C
=(1+x)log(1+x)-(1+x)+C
=(1+x){log(1+x)-1} + C

thats iofr 0 to 1
=(1+1){log(1+1)-1}
=2(log2-1)

this is the answer

Answer 4

use the rule that states that the integration of a negative logarithmic function is equal to the negative integration of a positive logarithmic function

Answer 5

By parts,

I log(1+x) = log(1+x)I(1) - I(d(log(1+x)I(1))

=xlog(1+x) - I(x/(1+x)) = xlog(1+x) - I(1-1/(1+x)) = xlog(1+x)-x-log(1+x)

Nox substitue x=0 and 1.

Answer 6

hey mate,

employ integration by parts,

int ( u(x)dv/dx ) = u(x)v(x) - int( v(x)du/dx )

For,
int(ln(x)) (assuming base 'e' - if not using the logarimthic conversion law to convert it to base 'e')

let u(x) = ln(x) --> du/dx = 1/x
dv/dx = 1 --> v(x) = x

Then,

int(ln(x)) = xln(x) - int(x.(1/x))
int(ln(x)) = xln(x) - int(1)
=xln(x) - x + C ; { C being the constant of integration}
=x(ln(x) - 1) + C

Answer 7

Hello ! , where are you.? It cannot be integrated !!!!