How do you integrate $ln(t+1)$ ?

Question

10150 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T00:46:10

$tln(t+1)- t + ln(t+1)+C$

The trick is to use integration by parts. That is we can re write the integral by making use of:

$int uv' dt = uv -int u'vdt$

We have:

$intln(t+1)dt$

Consider:

$int1.ln(t+1)$ dt

Set: $u = ln(t+1) -> u' = 1/(t+1)$

and $v' = 1 -> v =t$

We can now re write this integral as:

$tln(t+1) - intt/(t+1)dt$

$=tln(t+1) - int(t+1-1)/(t+1)dt$

$=tln(t+1) - int1 - 1/(t+1)dt$

Which will finally give us:

$tln(t+1) - t +ln(t+1) +C$

How do you integrate ln(t+1)ln(t+1)?