How do you integrate $(\frac{ln (x + 1)}{x^{2}}) d x$ $(ln(x+1)/(x^2)) dx$ ?

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Answer 1 · 2016-07-23T18:57:19

$= - (\frac{1}{x} + 1) ln (x + 1) + ln x + C$ $= - (1/x + 1) ln(x+1) + ln x +C$

$int \ (ln(x+1)/(x^2)) \ dx$

$= int \ ln(x+1) d/dx( -1/x) \ dx$

by IBP this becomes:

$=- 1/x ln(x+1) + int \ d/dx (ln(x+1)) * 1/x \ dx$

$=- 1/x ln(x+1) + int \ 1/(x(x+1)) \ dx$

so some partial fractions on this integral

$1/(x(x+1)) = A/ x + B/(x+1) = (A(x+1) + B x)/(x(x+1))$

$x = 0, 1 = A$
$x = -1, 1 = -B$

$implies - 1/x ln(x+1) + int \ 1/ x - 1/(x+1)\ dx$

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

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

How do you integrate (ln(x+1)/(x^2)) dx(ln(x+1)x2)dx(ln(x+1)/(x^2)) dx?