Skip to main content

`int sqrt(25x^2+4)/x^4 dx` Find the indefinite integral

To solve the indefinite integral, we follow `int f(x) dx = F(x) +C`


where:


`f(x)` as the integrand function


`F(x)` as the antiderivative of `f(x)`


`C` as the constant of integration.


For the given integral problem: `int sqrt(25x^2+4)/x^4dx` , we may apply integration by parts: `int u *dv = uv - int v *du` .


Let:


`u =sqrt(25x^2+4)`


Apply Law of Exponent: `sqrt(x) = x^(1/2)` , we get:  `u =(25x^2+4)^(1/2)`


To find the derivative of `u` , we may apply Power rule for derivative: `d/(dx) u^n= n* u^(n-1) * d/(dx)(u)`


`u' = 1/2 *(25x^2+4)^(1/2-1) * d/(dx) (25x^2+4)`


`u' = 1/2*(25x^2+4)^(-1/2)* (50xdx)`


`u' = 25x(25x^2+4)^(-1/2)dx`


Apply Law of exponent: `1/x^n =x^(-n)` .


`u' = (25x)/(25x^2+4)^(1/2)dx or(25x)/sqrt(25x^2+4)dx`


 Let: `v' = 1/x^4 dx`


To find the integral of `v'` , we apply  Law of exponent: `1/x^n =x^(-n)` and Power rule for integration:`int x^n dx = x^(n+1)/(n+1) +C` .


`v = int v'`


  `= int 1/x^4 dx`


  `= int x^(-4) dx`


  `= x^(-4+1)/(-4+1)`


  `= x^(-3)/(-3)`


  `= - 1/(3x^3)`


Apply the formula for integration by parts using the following values: `u =sqrt(25x^2+4)` , `u' =(25x)/sqrt(25x^2+4) dx` , `v'= 1/x^4 dx` and `v=- 1/(3x^3)` .


`int sqrt(25x^2+4)/x^4dx =sqrt(25x^2+4)*-( 1/(3x^3)) - int(25x)/sqrt(25x^2+4)*(- 1/(3x^3))dx`


                              `=-sqrt(25x^2+4)/(3x^3) - int -25/(3x^2sqrt(25x^2+4))dx`


To evaluate the integral part, we may apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .


`int -(25x)/(3x^3sqrt(25x^2+4))dx =-25/3int 1/(x^2sqrt(25x^2+4))dx`


The integral resembles one of the formulas from the integration table for rational function with roots. We follow:


`int (du)/(u^2sqrt(u^2+a^2)) = -sqrt(u^2+a^2)/(a^2x) +C`


For easier comparison, we may apply u-substitution by letting: `u^2 = 25x^2` or `(5x)^2` then `u=5x ` and `du = 5dx` or `(du)/5 = dx` . When we let `u^2 =25x^2` , it can be rearrange as `x^2=u^2/25` . Applying the values, the integral becomes:


`-25/3int 1/(x^2sqrt(25x^2+4))dx =-25/3int 1/((u^2/25)sqrt(u^2+4))*(du)/5`


                                        ` =-25/3int 25/(5u^2sqrt(u^2+4))du`


                                       ` = -125/3int 1/(u^2sqrt(u^2+4))dx`


By comparing "`u^2sqrt(u^2+a^2)` " with "`u^2sqrt(u^2+4)` ", we determine the corresponding value: `a^2=4` . Applying the aforementioned integration formula for rational function with roots, we get:


`-125/3int 1/(u^2sqrt(u^2+4))dx =-125/3* [-sqrt(u^2+4)/(4u)] +C`


                                      `=(125sqrt(u^2+4))/(12u) +C`


Plug-in `u^2= 25x^2 ` and `u =5x` on  `(125sqrt(u^2+4))/(12u) +C` , we get the indefinite integral:


`int -25/(3x^2sqrt(25x^2+4))dx=(125sqrt(25x^2+4))/(12*5x) +C`


                                       `=(25sqrt(25x^2+4))/(12x) +C` .


For the complete indefinite integral, we get:


`int sqrt(25x^2+4)/x^4dx =-sqrt(25x^2+4)/(3x^3) - int -25/(3x^2sqrt(25x^2+4))dx`


                             `=-sqrt(25x^2+4)/(3x^3)-(25sqrt(25x^2+4))/(12x) +C`

Comments

Post a Comment

Popular posts from this blog

Is there a word/phrase for "unperformant"?

As a software engineer, I need to sometimes describe a piece of code as something that lacks performance or was not written with performance in mind. Example: This kind of coding style leads to unmaintainable and unperformant code. Based on my Google searches, this isn't a real word. What is the correct way to describe this? EDIT My usage of "performance" here is in regard to speed and efficiency. For example, the better the performance of code the faster the application runs. My question and example target the negative definition, which is in reference to preventing inefficient coding practices. Answer This kind of coding style leads to unmaintainable and unperformant code. In my opinion, reads more easily as: This coding style leads to unmaintainable and poorly performing code. The key to well-written documentation and reports lies in ease of understanding. Adding poorly understood words such as performant decreases that ease. In addressing the use of such a poorly ...
Post a Comment
Read more

Is 'efficate' a word in English?

I routinely hear the word "efficate" being used. For example, "The most powerful way to efficate a change in the system is to participate." I do not find entries for this word in common English dictionaries, but I do not have an unabridged dictionary. I have checked the OED (I'm not sure if it is considered unabridged), and it has no entry for "efficate". It does have an entry for "efficiate", which is used in the same way. Wordnik has an entry for "efficate" with over 1800 hits, thus providing some evidence for the frequency of use. I personally like the word and find the meaning very clear and obvious when others use it. If it's not currently an "officially documented" word, perhaps its continued use will result in it being better documented.
Post a Comment
Read more