$$\int_{0}^{1} \frac{e^x}{x^6+x} dx$$
This how i approached the problem:
1st Step : Using partial fractions. $$e^x = A(x^5+1)+Bx^2+Cx$$
Now can i solve for $$Cx = e^x$$ and get $$C=e^{-ln(x)+x}$$
I want to understand if these are special kind of problem while, i dont think $$x^6+x)$$ is a polynomial that i could just factor. If someone has time and Energy to show me how it is done. Would highly appreciate it.
On
Here's an argument by partial fractions. We can write:
$$\frac{e^x}{x^6+x} = \frac{-x^4e^x}{x^5+1} +\frac{e^x-1}{x} + \frac{1}{x}$$
The first term on the right side is continuous and hence integrable on $[0,1]$. The second term can be made continuous by defining it as $1$ at $x=0$ and hence is also integrable on $[0,1]$.
Thus $$\int_0^1 \frac{e^x}{x^6+x}\,dx$$ converges if and only if $$\int_0^1\frac{dx}{x}$$ converges. It doesn't.
Notice that $$\frac{e^{x}}{x^{6}+x}\geq \frac{1}{x^{6}+x}\geq \frac{1}{2x}$$ for all $x\in [0,1]$
Since the integrand is non-negative on $[0,1]$, the integral diverges by simple comparison with $$\int_{0}^{1}{\frac{dx}{2x}}$$