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Our class finished integration, like taking the integral of $5x^2$ and $\sin{x}$.

For an extra credit problem, we were assigned:

1a. $\int x^{x^x}x^x\left(\ln^2(x)+\ln{x}+\frac1x\right) \,dx$

I had no clue how to do this, so I randomly took the derivative of $x^{x^x}$ using implicit differentiation and happened to get this exact integrand. However, how can I do this WITHOUT guess and check.

I want to do this by myself, so as an actual question, how would I go about integrating

$∫x^x(1+\ln{x})\,dx$, which $= x^x +C$.

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    'Guessing' is perfectly valid to be honest. Intuition is part of mathematics, and it's also something that is developed. That said, I don't think there's any implicit differentiation involved here. Implicit differentiation is for things like $x^2+y^2=c$ in which one (implicitly) assumes that $y$ is a function of $x$ and differentiaties both sides with respect to $x$. – 2017-02-06
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    you should put $x^x=t$ . Then $x^x(1 +log(x))dx=dt$. – 2017-02-06
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    You'd be surprised at how many integrals boil down to guesswork. The pros are just very good at guessing. Typically, if you have a difficult function, try setting it as $u$. Sometimes you might want to set a function inside of a difficult function as $u$ instead. Both work here. – 2017-02-06

1 Answers 1

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Start with $$\ln{x^{x^{x}}} = x^x \ln {x}$$

as this moves the power of $x$ down to become part of a product.

By differentiating both sides with respect to $x$ we may now use the product rule on one side:

$$\displaystyle\frac{1}{x^{x^x}}\displaystyle\frac{d}{dx}x^{x^x}=x^x \displaystyle\frac{1}{x} + \ln {x} \displaystyle\frac{d}{dx}x^x$$.

We can use the similar fact that

$$\ln{x^x} = x \ln {x}$$

and differentiate both sides again to get

$$\displaystyle\frac{1}{x^{x}}\displaystyle\frac{d}{dx}x^{x}=x \displaystyle\frac{1}{x} + \ln {x} \displaystyle\frac{d}{dx}x$$

which simplifies to

$$\displaystyle\frac{d}{dx}x^{x}= x^x(1 + \ln {x})$$.

Subtituting this into our first differentiation we have

$$\displaystyle\frac{1}{x^{x^x}}\displaystyle\frac{d}{dx}x^{x^x}=x^x \displaystyle\frac{1}{x} + x^x \ln {x} (1 + \ln {x})$$

which can then be rearranged to form:

$$\displaystyle\frac{d}{dx}x^{x^x}=x^{x^x} x^x (\ln^2{x}+\ln x + \displaystyle\frac{1}{x})$$

Side note Often the best ways to solve complicated integrals are:

a) Employing various techniques (integration by parts, substitution, recurrence relations etc.)

b) Rearranging the integrand to look more palatable/like a known derivative

c) Observing a pattern in the expression