How do you find the integral #(ln(x+1)/(x^2)) dx#?

1 Answer
bp
Aug 6, 2015

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

Explanation:

Integrate the function by parts, being a product of functions #1/x^2# and ln (x+1)

The integral would be #ln (x+1) (-1/x)# +#int 1/(x(x+1)) dx#

# (-1/x) ln(x+1) + int (1/x - 1/(x+1)) dx#

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

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