What is the antiderivative of $(ln(x+1)/(x^2))$ ?

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Answer 1 · 2016-01-09T01:09:11

$I = -ln(x+1)/x + ln|x| - ln|x+1| + c$

$I = intln(x+1)/x^2dx$

Let's say $u = ln(x+1)$ so $du = 1/(x+1)$ , and $dv = 1/x^2$ so $v = -1/x$

$I = -ln(x+1)/x + intdx/(x(x+1))$

The latter integral can only be solved with parcial fractions, so we assume there is a sum of fractions

$a/x + b/(x+1) = 1/(x(x+1))$

Where $a$ and $b$ are constants, that is also to say

$(a(x+1) + bx)/(x(x+1)) = 1/(x(x+1))$

Note that it means, that, for any value of $x$

$a(x+1) + bx = 1$

So if we assume $x = 0$ ,

$a(0+1) + b*0 = 1$
$a = 1$

And if we assume $x = -1$

$a(-1+1) -b = 1$
$-b = 1$
$b = -1$

So, back to the first integral we have

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

From there, it's an easy integral

$I = -ln(x+1)/x + ln|x| - ln|x+1| + c$

What is the antiderivative of (ln(x+1)/(x^2)) (ln(x+1)/(x^2))?