what is the integral of (ln(x+1)/(x^2)) dx?

Question

it's very easy, think a little!!!!

Answer 1

this problem can be solved by using integration by parts

int(u)dv =uv -int(v)du

putting u = ln(x+1) and dv/dx =1/x^2 >>>>>> v= -1/x

and du/dx = 1/(1+x)

int(ln(x+1)/x^2 =ln(x+1).-1/x +A -int(-1/x.1/(x+1))dx

= -ln(x+1)x + A + int(1/x(x+1))dx

using partial fractions,

1/x(x+1) = 1/x - 1/(x+1)

so, int (1/x(x+1))dx = int(1/x)dx-int(1/(x+1))dx

=ln x-ln(x+1)+B say

let A+B=C say,a constant

therefore,

int(ln(x+1)/x^2)dx = -ln(x+1)/x - ln(x+1) +ln x +C

= -(x+1)/x(ln(x+1)) +ln x +C

= ln x - (x+1)(ln(x+1))/x + C

note that this answer is subject to the restriction that x cannot be equal to 0 or -1
when x=0 or-1,the above answer is a singularity- when you are taking the log of less than 0-,the log is a complex number

i checked this answer by differentiating it by the quotient rule
and produced the original integral-it is therefore correct

Answer 2

4

Answer 3

Integral of ln(x+1)/x^2 with respect to x is:

ln(x) - (x+1)*ln(x+1)/x + constant

Answer 4

∫ [ ln ( x+1 ) / x² ] dx =
Ln[x] - ( (1 + x) * ln[1 + x] ) / x

Answer 5

(ln(x+1)/(x^2)) dx
=(ln(x+1)-ln(x^2))dx
=(ln(x+1)-2ln(x))dx
=1/(x+1)-2/x+C
surely there's something i'm missing...

Answer 6

log(x) - ((x +1)log(x+1) )/x

When you want to integrate things. Use Wolfram's integrator.

It saved my life when i took calculus 1,2,3 and diff equ.

Answer 7

now enter plis you ta tei jajajajaja que preguntas guapo

Answer 8

No guapo yo con tremendas grocerias no me llevo, jajaja, saludos

Answer 9

If that avatar is a pic of you , you are hot !

Answer 10

(-((x+1)-x*ln(x)))/x