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

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
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Recognitions: Gold Member Science Advisor 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.
 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

Recognitions:
Homework Help

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

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

 Similar discussions for: Differentiation [ (7x^3+4)^(1/x) ] / Integration [ x^x(1+ln(x)) ] Thread Forum Replies General Math 9 Calculus 5 Calculus & Beyond Homework 9 Linear & Abstract Algebra 5 Introductory Physics Homework 5