Evaluate $\int x^{x^n+n-1}(x \ln x +1)\mathrm dx$

Question

$$\int x^{x^n+n-1}(x \ln x +1) \mathrm dx$$

I tried integration by part, I had no result. can someone help?

Answer 1

This is general form of integral:

$$\int x^x(\ln x+1)dx= x^x$$

Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;

$$y'=x^{x^n+n-1}(n\ln x+1)=x^{x^n}.x^{n-1}(n\ln x+1)$$

⇒ $$x^{n-1}(n\ln x+1)=n x^{n-1} \ln x +\frac{x^n}{x}=(x^n \ln x)'=(\ln x^{x^n})'$$

⇒ $$y'=x^{x^n+n-1}(n\ln x+1)=x^{x^n}.(\ln x^{x^n})'=$$

Now we compare this with:

$$\ y=f(x)⇒\ln y =\ln f(x)⇒\frac{y'}{y}=[\ln f(x)]'$$

and conclude that:

$$ y=x^{x^n}$$

Answer 2

I think the problem is $$\int x^{x^n+n-1}(n \ln x +1) dx$$

If $y=x^{x^n}$

$\ln y=x^n\ln x$

$$\frac{1}{y}\dfrac{dy}{dx}=nx^{n-1}\ln x+x^{n-1}$$

$$\implies\dfrac{dy}{dx}=x^{x^n+n-1}(n\ln x+1)$$