Indefinite Integral

Question

Can someone help me solve for this indefinite integral?:

 

(u^2 + u)/(2u^3 + 3u^2 + 1)^(2/3)

 

Thank you!

Answer 1

We need to rewrite the integral so that it is in the form of a product.

 

∫(u² + u)(2u³ + 3u² + 1)^-2/3 du

 

 

We can use the substitution method.

 

Let      x = 2u³ + 3u² + 1

 

        dx = (6u² + 6u)du

 

        dx = 6(u² + u)du

 

       (1/6)dx = (u² + u)du

 

 

Using these substituted terms, we can write an equivalent integral that is easier to evaluate.

 

∫(1/6)x^-2/3dx =

 

(1/6)∫x^-2/3 dx =

 

(1/6)(3)x^1/3 + C =

 

(1/2)x^1/3 + C

 

 

Now we substitute the value of x back into the integral so that it is in terms of u.

 

(1/2)(2u³ + 3u² + 1)^1/3 + C

 

 

To check, if we take the derivative of the solution, we should get back the original expression that we need to integrate.

Answer 2

∫[u²+u]/[2u³+3u²+1]^2/3du

 

   Let w = 2u³+3u²+1  then dw = (6u²+6u)du

 

                                     So, (1/6)dw = (u²+u)du

 

The given integral can then be rewritten as (1/6)∫w^-2/3dw

 

                                                 = (1/6)(3w^1/3) + C

 

                                                 = (1/2)(2u³+3u²+1)^1/3 + C