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

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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#