Question

How do you evaluate the definite integral `int (1 + ln x)^2 / x` from 1 to e?

Answer 1

We can see that `(1+lnx)'=1/x` hence

#int_1^e (1+lnx)'*(1+lnx)^2dx=1/3 int_1^e [(1+lnx)^3]'dx= 1/3[ln(1+x)^3]_1^e=1/3*[ln(1+e)^3-ln2^3]#

Note `()'=d/dx`

Answer 2

`7/3`

Explanation:

Let `u=lnx`.

This implies that `du=1/xdx`, which we see we have, since there is an `x` in the denominator.

Thus, we have

`int_1^e(1+lnx)^2/xdx=int_0^1(1+u)^2du`

Notice that the bounds also changed, since we substituted in `u`. The new bounds were found by plugging the old bounds into `u=lnx`, as follows:

`u(1)=ln(1)=0`
`u(e)=ln(e)=1`

Expand `(1+u)^2`:

`=int_0^1(1+2u+u^2)du`

We can integrate `int_0^1(1+2u+u^2)du` fairly easily using `intu^ndu=u^(n+1)/(n+1)+C`:

`=[u+u^2+u^3/3]_0^1=(1+1^2+1^3/3)-(0+0^2+0^3/3)`

`=7/3`