Expert Answer to "How do you integrate (exx)dx ?"

Ashley Bell

Answered question

2021-12-13

How do you integrate

(\frac{e^{x}}{x}) d x

Laura Worden

Beginner2021-12-14Added 45 answers

This is sometimes called the exponential integral:

\int \frac{e^{x}}{x} d x = E i (x) + C

But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for

e^{x}

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}

Then:

\frac{e^{x}}{x} = \frac{1}{x} + 1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \dots = \frac{1}{x} + \sum_{n = 0}^{\infty} \frac{x^{n}}{(n + 1)!}

So the antiderivative will be:

\int \frac{e^{x}}{x} d x = \int (\frac{1}{x} + 1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \dots) d x

= \ln (| x |) + x + \frac{x^{2}}{2 \cdot 2!} + \frac{x^{3}}{3 \cdot 3!} + \dots + C

\int \frac{e^{x}}{x} d x = \ln (| x |) + \sum_{n = 1}^{\infty} \frac{x^{n}}{n \cdot n!} + C

Joseph Lewis

Beginner2021-12-15Added 43 answers

Answer:

e^{x} = \frac{x^{0}}{0} + \frac{x^{1}}{1} + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots

= 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots

\int \frac{e^{x}}{x} d x

= \int \frac{1}{x} (1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots) d x

= \int (\frac{1}{x} + 1 + \frac{x}{2} + \frac{x^{2}}{3} + \dots) d x

= \log | x | + x + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{3^{3}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?