Question

∫▒1/x.e^xdx

Answer 1

Answer on Question#39060 – Math – Integral Calculus

Evaluate: $\int \frac{1}{x\,e^x} \, dx$.

Solution.

Integration by parts :

If $u = u(x)$, $v = v(x)$, and the differentials $du = u'(x)\,dx$ and $dv = v'(x)\,dx$, then integration by parts states that

$$\int u(x) v'(x)\,dx = u(x) v(x) - \int u'(x) v(x)\,dx$$

or more compactly:

$$\int u\,dv = uv - \int v\,du.$$

So,

$$\begin{aligned} \int \frac{1}{x\,e^x} \, dx &= \int \frac{e^{-x}}{x} \, dx = - \int \frac{d(e^{-x})}{x} = - \left( \frac{1}{x \cdot e^x} - \int \frac{-1}{x^2\,e^x} \, dx \right) = \\ &= \frac{-1}{x \cdot e^x} + \int \frac{1}{x^2} \, d(e^{-x}) = \frac{-1}{x \cdot e^x} + \frac{1}{x^2 \cdot e^x} - \int \frac{-2}{x^3\,e^x} \, dx = \\ &= \frac{-1}{x \cdot e^x} + \frac{1}{x^2 \cdot e^x} - \int \frac{2}{x^3} \, d(e^{-x}) = \\ &= \frac{-1}{x \cdot e^x} + \frac{1}{x^2 \cdot e^x} - \left( \frac{2}{x^3 \cdot e^x} - \int \frac{-2 \cdot 3}{x^4\,e^x} \, dx \right) = \\ &= \frac{-1}{x \cdot e^x} + \frac{1}{x^2 \cdot e^x} - \frac{2}{x^3 \cdot e^x} + \int \frac{6}{x^4} \, d(e^{-x}) = \dots = \\ &= e^{-x} \left( -\frac{1}{x} + \left( \frac{1}{x} \right)^2 - \frac{2}{x^3} + \frac{6}{x^4} - \frac{24}{x^5} + O\left( \left( \frac{1}{x} \right)^6 \right) \right) \end{aligned}$$ $$\int \frac{1}{x\,e^x} \, dx = Ei(-x) + \text{const},$$

where $Ei(x)$ is the exponential integral.