Question

How would you integrate 1+lnx ?

Answer 1

lnx isn't super obvious just by looking at it.

It will require integration by parts.

Answer 2

Get the 1 out of the way so we can ignore it.\[\large\rm \int\limits 1+\ln x~dx\quad=\quad x+\int\limits \ln x~dx\]

Answer 3

\[\large\rm u=\ln x\qquad\qquad\qquad dv=dx\]Integrating lnx would be... hmm weird. That's what we're trying to get around anyway. So we want to differentiate our log instead. Hence, it is our u.

Answer 4

What's dv?

Answer 5

It's the invisible 1 next to the ln(x).

ln(x) = 1*ln(x)

We're integrating the 1.

Answer 6

oh ok

Answer 7

It's better to get comfortable with differentials though.
dv = dx

If that's confusing though, and you'd rather look at the function stuff,
then yes, it's the 1.

Answer 8

\[\large\rm u=\ln x\qquad\qquad\qquad dv=dx\]So we get our other pieces that we need,\[\large\rm du=\frac1x~dx\qquad\qquad\qquad v=x\]

Answer 9

ok looks good

Answer 10

So umm.. one of things you might notice is...
the log turns into a power of x when it's differentiated.
That will mix nicely with our new v showing up.

Answer 11

So we get something like this from our parts, yes?\[\large\rm \int\limits \ln x~dx\quad=\quad x \ln x-\int\limits x\cdot\frac{1}{x}~dx\]Do you see how that new integral that shows up is much easier to deal with now?

Answer 12

ohh ok i see

Answer 13

\[\large\rm \int\limits\limits \ln x~dx\quad=\quad x \ln x-\int\limits dx\]

Answer 14

\[\large\rm \int\limits\limits\limits \ln x~dx\quad=\quad x \ln x-x+c\]

Answer 15

And then don't forget about the +1 that we integrated earlier on, right?

Answer 16

oh yeah so that would then cancel out this x right?

Answer 17

\[\large\rm \int\limits(1+\ln x)~dx\quad=\quad x \ln x+c\]Yayyy good job \c:/

Answer 18

thank you!!! This helped a lot!!!!

Answer 19

yay team