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

1 Answer
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