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

1. Apr 20, 2007

1. The problem statement, all variables and given/known data
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

2. Apr 20, 2007

### HallsofIvy

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

3. Apr 20, 2007

### 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

Last edited: Apr 20, 2007
4. Apr 20, 2007

### 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.

5. Apr 20, 2007