12 Days of Christmas Integrals: ∫ x^n ∙ e^x dx; n is a positive integer

12 Days of Christmas Integrals:  ∫ x^n ∙ e^x dx;  n is a positive integer 

On the Twelfth day of Christmas Integrals, the integral featured today is...

∫ x^n ∙ e^x dx;  n is a positive integer 

n = 1, 2, 3, 4, ...

Each term is found by repeating the Integration by Parts method.  Today, we will find a (somewhat) closed formula that applies to any positive integer n.  

∫ x^n ∙ e^x dx

u = x^n

du = n ∙ x^(n-1) dx

dv = e^x dx

v = e^x

= x^n ∙ e^x - ∫ n ∙ x^(n-1) ∙ e^x dx

u = n ∙ x^(n-1)

du = n ∙ (n-1) ∙ x^(n-2) dx

dv = e^x dx

v = e^x

= x^n ∙ e^x - n ∙ x^(n-1) ∙ e^x + ∫ n ∙ (n-1) ∙ x^(n-2) ∙ e^x dx

u = n ∙ x^(n-1)

du = n ∙ (n-1) ∙ x^(n-2) dx

dv = e^x dx

v = e^x

and so on until...

u = n ∙ (n - 1) ∙ ... ∙ 2 ∙ x

du = n! dx

dv = e^x dx

v = e^x

with the closed form:

= e^x ∙ [x^n - n ∙ x^(n-1) + n ∙ (n-1) ∙ x^(n-2) - n ∙ (n-1) ∙ (n-2) ∙ x^(n-3) + ...  

   + (-1)^(n mod 2) ∙ n!]

Examples: 

n = 2

∫ x^2 ∙ e^x dx = e^x ∙ (x^2 - 2 ∙ x + 2)

n = 3

∫ x^3 ∙ e^x dx = e^x ∙ (x^3 - 3 ∙ x^2 + 6 ∙ x - 6)

n = 4

∫ x^4 ∙ e^x dx = e^x ∙ (x^4 - 4 ∙ x^3 + 12 ∙ x^2 - 24 ∙ x + 24)

What do you think of the integral series?  Is this worth a sequel? 

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a,b are constants

 12 Days of Christmas Integrals:  ∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a, b are constants

On the Eleventh day of Christmas Integrals, the integral featured today is...

∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a, b are constants

A rather simple integral:

∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx

= a ∙ b ∙ ∫ 1÷ (a^2 ∙ x^2 + b^2) dx

= a ∙ b ∙ ∫ 1÷ (b^2 ∙ (a^2/b^2 ∙ x^2 + 1)) dx

= a/b ∙ ∫ 1÷ (a^2/b^2 ∙ x^2 + 1) dx

= ∫ (a/b) ÷ ((a/b)^2 ∙ x^2 + 1) dx

= arctan(a/b ∙ x) + C

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx

 12 Days of Christmas Integrals:  ∫ x^2 ∙ exp(x)^2 dx   

On the Tenth day of Christmas Integrals, the integral featured today is...

∫ x^2 ∙ exp(x)^2 dx   

exp(x) = e^x

Start with the substitution:

u = exp(x)

du = exp(x) dx

1 ÷ exp(x) du = dx

1 ÷ u du = dx 

u^2 = exp(x)^2

ln(u) = x

ln(u)^2 = x^2

Then:

∫ x^2 ∙ exp(x)^2 dx   

= ∫ ln(u)^2 ∙ u^2 ∙ 1/u du

= ∫ u ∙ ln(u)^2 du

Note:  We did this integral on the Seventh Day of 12 Days of Christmas integrals (December 31, 2021).  Check out that post for details!

= u^2/2 ∙ ln(u)^2 - u^2/2 ∙ ln(u) + u^4/4 + C

Recall u = exp(x):

= 1/2 ∙ (exp(x))^2  ∙ x^2 - 1/2 ∙ (exp(x))^2 ∙ x + 1/4 ∙ (exp(x))^2 + C 

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant

 12 Days of Christmas Integrals:  ∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant

On the Ninth day of Christmas Integrals, the integral featured today is...

∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant  

Multiply by (√(x+a) + √(x)) ÷ (√(x+a) + √(x)):

= ∫ 1 ÷ (√(x + a) - √(x)) ∙ (√(x+a) + √(x)) ÷ (√(x+a) + √(x)) dx

= ∫ (√(x+a) + √(x)) ÷ (x + a - x) dx

= ∫ (√(x+a) + √(x)) ÷ (a) dx

= ∫ √(x + a) ÷ a + √(x) ÷ a dx

= 2/3 ∙ (x + a)^(3/2) + 2/3 ∙ x^(3/2) ÷ a

= 2 ÷ (3 ∙ a) ∙ ((x + a)^(3/2) + x^(3/2))

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

 12 Days of Christmas Integrals:  ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

HAPPY NEW YEAR!!!!   HERE'S TO A WONDERFUL 2022!  May we make our future and present bright!  (We all could use a bright year, given how the last few years went)

On the Eighth day of Christmas Integrals, the integral featured today is...

∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

By division:

= ∫ 1 + (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) dx

Find the partial fraction of (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) 

(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = A ÷ (x - 2) + B ÷ (x - 4)

2 ∙ x - 5 = A ∙ (x - 4) + B ∙ (x - 2)

2 ∙ x - 5 = A ∙ x - 4 ∙ A + B ∙ x - 2 ∙ B

which implies that:

-5 = -4 ∙ A - 2 ∙ B

2 = A + B

Solving for A and B:

A  = 1/2 and B = 3/2 

(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4)

= ∫ 1 + (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4) dx

= x + 1/2 ∙ ln|x - 2| + 3/2 ∙ ln|x - 4| + C

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ x ∙ (ln(x))^2 dx

12 Days of Christmas Integrals:  ∫ x ∙ (ln(x))^2 dx

NEW YEARS EVE!!!!

On the Seventh day of Christmas Integrals, the integral featured today is...

∫ x ∙ (ln(x))^2 dx

Sounds like a job for integration by parts!

∫ x ∙ (ln(x))^2 dx

u = (ln(x))^2 

du = 2 ∙ ln(x) ∙ 1/x dx

dv = x dx

v = x^2/2

= x^2/2 ∙ (ln(x))^2 - ∫ 2 ∙ ln(x) ∙ 1/x  ∙ x^2/2 dx

= x^2/2 ∙ (ln(x))^2 - ∫ x ∙ ln(x) dx

u  = ln(x)

du = 1/x dx

dv = x dx

v = x^2/2

= x^2/2 ∙ (ln(x))^2 - x^2/2 ∙ ln(x) + ∫ x/2 dx

= x^2/2 ∙ (ln(x))^2 - x^2/2 ∙ ln(x) + x^2/4 + C

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ cos (√x) dx

12 Days of Christmas Integrals:  ∫ cos (√x) dx

On the Sixth day of Christmas Integrals, the integral featured today is...

∫ cos (√x) dx

Handling this integral will require two integral methods.  First substitution:

Let u = √x = x^(1/2)

Then:

du = 1/2 ∙ x^(-1/2) dx

2 ∙ x^(1/2) du = dx

2 ∙ u du = dx

∫ cos(√x) dx

= ∫ cos(x^(1/2)) dx

= ∫ 2 ∙ u ∙ cos(u) du

At this point, we now apply Integration by Parts:

w = 2 ∙ u

dw = 2 du

dv = cos(u) du

v = sin(u)

= 2 ∙ u ∙ sin(u) - ∫ 2 ∙ sin(u) du

= 2 ∙ u ∙ sin(u) +  2 ∙ cos(u) + C

Recall u = x^(1/2):

= 2 ∙ x^(1/2) ∙ sin(x^(1/2)) + 2 ∙ cos(x^(1/2)) + C

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

12 Days of Christmas Integrals:  ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

On the Fifth day of Christmas Integrals, the integral featured today is...

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

I am going to use two approaches and show why both approaches are valid.  Remember we are working indefinite integrals.  

Substitution Method

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

u = x^2 + 1 

du = 2 ∙ x dx

u^2 = (x^2 + 1)^2

∫ u^2 du

= u^3 ÷ 3 + C

= (x^2 + 1)^3 ÷ 3 + C

= (x^6 + 3 ∙ x^4 + 3 ∙ x^2 + 1) ÷ 3 + C

= (x^6 + 3 ∙ x^4 + 3 ∙ x^2) ÷ 3 + 1 ÷ 3 + C

= x^6 ÷ 3 +  x^4 +  x^2 + 1 ÷ 3 + C

Algebraic Method

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

= ∫ 2 ∙ x ∙ (x^4 + 2 ∙ x^2 + 1) dx

= ∫ 2 ∙ x^5 + 4 ∙ x^3 + 2 ∙ x dx

= 2/6 ∙ x^6 + 4/4 ∙ x^4 + 2/2 ∙ x^2 + C

= x^6 ÷ 3 + x^4 + x^2 + C

Why are the two methods acceptable?

In indefinite integration, a "+ C" is added to show that the solutions to indefinite integrals are a class of functions.   C is an arbitrary numerical constant.

d/dx[  f(x) + C ] =   d/dx [ f(x) ] + 0  = d/dx [ f(x) ]

Since C is a numerical constant,  C + 1/3 is also a numerical constant.  

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ x^2 ÷ (1 + x^2) dx

 12 Days of Christmas Integrals:  ∫ x^2 ÷ (1 + x^2) dx

On the Fourth day of Christmas Integrals, the integral featured today is...

∫ x^2 ÷ (1 + x^2) dx

Time for a little algebraic manipulation:  add and subtract 1 ÷ (1 + x^2) 

= ∫ x^2 ÷ (1 + x^2) + 1 ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ (x^2 + 1) ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ (1 + x^2) ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ 1 - 1 ÷ (1 + x^2)   dx

= x - arctan(x) + C

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ (sin(x))^2 dx

 12 Days of Christmas Integrals:  ∫ (sin(x))^2 dx

On the Third day of Christmas Integrals, the integral featured today is...

∫ (sin(x))^2 dx

This calls for the trig identity:

(sin(x))^2 = 1/2 - 1/2 ∙ cos(2 ∙ x)

With any operations in calculus involving trigonometric functions, the angle units are assumed to be in radians.

∫ (sin(x))^2 dx 

= ∫ 1/2 - 1/2 ∙ cos(2 ∙ x) dx

= 1/2 ∙ x - 1/4 ∙ sin(2 ∙ x) + C

= 1/2 ∙ x - 1/2 ∙ sin(x) ∙ cos(x) + C

Simple as that.   We will get meatier integrals as the days wear on.  

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

 12 Days of Christmas Integrals:  ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

On the Second day of Christmas Integrals, the integral featured today is...

∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

This integral can be approached by using the substitution method:  

Let u = arctan(x^2).  Then:

du = 1 ÷ (1 + x^4) ∙  d/dx(x^2)

du = 1 ÷ (1 + x^4) ∙  (2  ∙ x) dx

1/2 du = x ÷ (1 + x^4) dx

Then:

∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

= ∫ u ∙ 1/2 du

= 1/2 ∙ ∫ u du

 = u^4/4 + C

Substitute back:

= (arctan(x^2))^2 / 4 + C

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

12 Days of Christmas Integrals: ∫ π ∙ cos(x) ∙ sin(x) dx

 12 Days of Christmas Integrals:  ∫ π ∙ cos(x) ∙ sin(x) dx

In the spirit of the Christmas Holiday, I am presenting you with the 12 Days of Christmas Integrals!

On the First day of Christmas Integrals, the integral featured today is...

∫ π ∙ cos(x) ∙ sin(x) dx

Let's start off with using the trigonometric identity

sin(2 ∙ x) = 2 ∙ cos(x) ∙ sin(x)

1/2 ∙ sin(2 ∙x) = cos(x) ∙ sin(x)

Then:

∫ π ∙ cos(x) ∙ sin(x) dx

= π ∙ ∫ 1/2 ∙ sin(2 ∙ x) dx

= π/2 ∙ ∫ sin(2 ∙ x) dx

Multiply by both 1/2 and 2:

= π/4 ∙ ∫ 2 ∙ sin(2 ∙ x) dx

= -π/4 ∙ cos(2 ∙ x) + C 

[ ∫ sin x dx = -cos x + C,  angles are in radians ]

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.