Question

Question #70eda

Answer 1

`int_2^6sqrt(x)logxdx=`

`2/(3ln10)((ln6-2/3)sqrt(216)+(2/3-ln2)sqrt8)`

Explanation:

We know that :

`logx=lnx/ln10`, so let's use this.

`int_2^6sqrt(x)logxdx=1/ln10int_2^6x^(1/2)lnxdx=`

we use integration by parts, https://en.wikipedia.org/wiki/Integration_by_parts

`1/ln10([2/3x^(3/2)lnx]_2^6-int_2^6(2x^(3/2))/(3x)dx)=`

`1/ln10((2/3 6^(3/2)ln6-2/3 2^(3/2)ln2)-2/3int_2^6x^(1/2)dx)=`

`2/(3ln10)(6^(3/2)ln6-2^(3/2)ln2-2/3[x^(3/2)]_2^6)=`

`2/(3ln10)(6^(3/2)ln6-2^(3/2)ln2-2/3 6^(3/2)+2/3 2^(3/2))=`

`2/(3ln10)((ln6-2/3)sqrt(216)+(2/3-ln2)sqrt8)`

Answer 2

Integrals of the form `x^r logx` belong in your mathematical toolbox (or cookbook -- choose your image).

Explanation:

I assume that you are using `logx` for the natural logarithm. If you want the common log (the base 10 log) rewrite `log_10x = lnx/ln10`. This will result in a factor of `1/ln10` in front of the answer I give.

`int x^r log(x) dx`

Integrate by parts with `u = logx` and `dv = x^r`. Rather than continuing with the general case, let's do the one you asked about.

`int x^(1/2) log(x) dx`

Integrate by parts with `u = logx` and `dv = x^(1/2) dx` so that

`du = 1/x dx` and `v = 2/3x^(3/2)`

`uv-int vdu = 2/3x^(3/2)logx - int 2/3x^(3/2) 1/x dx`

`= 2/3x^(3/2)logx - 2/3 int x^(3/2) x^-1 dx`

`= 2/3x^(3/2)logx - 2/3 int x^(1/2) dx`

`= 2/3x^(3/2)logx - 2/3 (2/3x^(3/2))`

`= 2/3x^(3/2)logx - 4/9 x^(3/2)`

Finish by evaluating from `2` to `6`.

General case

The integral `int vdu` will always simplify to a constant time the integral you just did to find `v`.

Bonus Example 1
`int x^3 log(x) dx`

Integrate by parts with `u = logx` and `dv = x^3 dx` so that

`du = 1/x dx` and `v = 1/4x^4`

`uv-intvdu = 1/4x^4logx-int 1/4x^4 1/x dx`

`= 1/4x^4logx-1/4 int x^3 dx` (We integrated `x^3` in step 1.)

Bonus Example 2

`int x^-7 log(x) dx`

Integrate by parts with `u = logx` and `dv = x^-7 dx` so that

`du = 1/x dx` and `v = -1/6 x^-6`

`uv-intvdu = -1/6x^-6 logx - int -1/6 x^-6 1/x dx`

`= -1/6x^-6logx+ 1/6 int x^-7 dx` (We integrated `x^-7` in step 1.)

Two more bonuses
One:
We use the same method to find `int logx dx`.

`u = logx` and `dv = 1dx`

Two:

If you like, you can work out the general rule for

`int x^rlogx dx = 1/(r+1) x^(r+1)logx - 1/(r+1)^2 x^(r+1) +C`

Answer 3

`2/3(6sqrt6ln6-2sqrt2ln2)-4/9(6sqrt6-2sqrt2).`

Explanation:

Let, `I=int_2^6 sqrtx*lnxdx=int_2^6x^(1/2)*lnxdx.`

We use the following Rule of Integration by Parts (IBP) :

IBP : `int_a^bu*vdx=[uintvdx]_a^b-int_a^b{(du)/dxintvdx}dx.`

We take : `u=lnx, and, v=x^(1/2).`

`:. (du)/dx=1/x, and, intvdx=x^(1/2+1)/(1/2+1)=2/3x^(3/2).`

`:. I=[2/3x^(3/2)*lnx]_2^6-int_2^6{1/x*2/3x^(3/2)}dx,`

`=2/3[6^(3/2)ln6-2^(3/2)ln2]-2/3int_2^6x^(1/2)dx,`

`=2/3(6sqrt6ln6-2sqrt2ln2)-2/3[x^(3/2)/(3/2)]_2^6,`

`rArr I=2/3(6sqrt6ln6-2sqrt2ln2)-4/9(6sqrt6-2sqrt2).`