Question #70eda

3 Answers
Jul 22, 2017

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)

Jul 22, 2017

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

Jul 22, 2017

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).