Question

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

Answer 1

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

Explanation:

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`