integrate from 0 to 2 ln (x+1)...... I used integration by parts but I am stuck after setting up the new integral

Question

terenzreignz · Answer

$$\large \int_0^2 \ln(x+1)dx$$
This it?

anonymous · Answer

yes

terenzreignz · Answer

Well, actually, there was no need for integration by parts...
Can you evaluate this antiderivative?
$$\large \int \ln \ x  \ dx$$

anonymous · Answer

There's no need for integrating by parts... let u = x + 1. Now we have $\int_1^3\ln u \ \mathrm{d}u$.

anonymous · Answer

the homework assignment asks to use parts

anonymous · Answer

... though, if you don't know $\int\ln x\ \mathrm{d}x$ then you can determine it using integration by parts. Let $u=\ln x$, $\mathrm{d}v=\mathrm{d}x$ and thus $\mathrm{d}u=\frac1x\mathrm{d}x$, $v=x$.

Now we have the integral $\int u\ \mathrm{d}v=uv-\int v\ \mathrm{d}u$. Can you reduce now?

anonymous · Answer

ok ill give it a shot thanks for your help

terenzreignz · Answer

Must have been confused.... yeah, integration by parts was key here, my bad...

anonymous · Answer

Not really, @terenzreignz. Typically you'd already know that integral.

terenzreignz · Answer

Still, no way to derive it other than through integration by parts.  There's no escaping it.
And no, we were not taught that integral before we learned integration by parts, just like we were not taught the integral of tan x before we derived the integral of 1/x, etc...

anonymous · Answer

Yes I agree with you terenzreignz.

anonymous · Answer

@ oldrin.bataku
 It seemed too easy to just do a simple u sub, thus being in chap 7 integration by parts i kept forcing myself to use that method to solve the problem. Yet I am still stuck after setting up the equation for parts.