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

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marizi eNotes educator| Certified Educator

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`

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