Differentiation [ (7x^3+4)^(1/x) ] / Integration [ x^x(1+ln(x)) ]

[unique_pavadrin]

Homework Statement

Differentiate the following expression leaving in the simplest form:
$$\left( {7x^3 + 4} \right)^{\frac{1}{x}}$$

Integrate the following leaving in the simplest form:
$$x^x \left( {1 + \ln x} \right)$$

2. The attempt at a solution
Here is my worked solution for the differentiation problem:

$$\[ \begin{array}{l} \frac{d}{{dx}}\left[ {\left( {7x^3 + 4} \right)^{\frac{1}{x}} } \right] \\ y = \left( {7x^3 + 4} \right)^{\frac{1}{x}} \\ \ln y = \left( {\frac{1}{x}} \right)\ln \left( {7x^3 + 4} \right) \\ = \frac{{\ln \left( {7x^3 + 4} \right)}}{x} \\ \frac{d}{{dx}}\left[ {\ln \left( {7x^3 + 4} \right)} \right] = \frac{1}{{7x^3 + 4}}.\frac{d}{{dx}}\left[ {7x^3 + 4} \right] \\ = \frac{{21x^2 }}{{7x^3 + 4}} \\ \frac{d}{{dx}}\left[ {\frac{{\ln \left( {7x^3 + 4} \right)}}{x}} \right] = \frac{{x\left( {\frac{{21x^2 }}{{7x^3 + 4}}} \right) + 1\left( {\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 }} \\ = \frac{{21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)}}{{x^2 }} \\ \\ \frac{1}{y} = \\ y' = y.\frac{{21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)}}{{x^2 \left( {7x^3 + 4} \right)}} \\ = \frac{{\left( {7x^3 + 4} \right)^{\frac{1}{x}} \left( {21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 \left( {7x^3 + 4} \right)}} \\ = \frac{{\left( {7x^3 + 4} \right)^{\frac{1}{x} - 1} \left( {21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 }} \\ \end{array} \]$$

______________________________________​

For the integration problem I am not quite certain on how to integrate the expression given. I know from previous experience that the expression x^x when differentiated will give x^x(1+ln(x)).

______________________________________​

many thanks for all suggestions and help
unique_pavadrin

 

[HallsofIvy]

For the integration, my first thought was to make a substitution like u= x^x. Of course, to do that I would need a derivative for x^x. Find the derivative of x^x and think you will realize that the integral is surprizingly easy.

 

[lalbatros]

I got this result, with a small difference:

(4 + 7*x^3)**(-1 + 1/x)*(21*x^3 - (4+7*x^3)*Log(4 + 7*x^3)))/x^2

The mistake appeared probably in the 7th line of your calculation.
You probably forgot the minus sign from the rule:

(a/b)' = (a'b-ab')/b²

 

[Gib Z]

Since from previous experience you know the integral is x^x, and x^x happens to appear in that integrand, the easiest way out will always be the substitution which gives the whole integrand! Same goes for any integral like that.

 

[unique_pavadrin]

thank you for the reply and thank you lalbatros for point out my stupid mistake