Question

Evaluate the indefinite integral as an infinite series. 44.\(\int {\frac{{{e^x} - 1}}{x}dx} \)

Short answer

\(\int {\frac{{{e^x} - 1}}{x}dx = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{(n)(n)!}} + C} } \)

Step-by-step solution

Step 1: Simplify and rewrite the given expression

We know that

\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \)

This can be written as

\(\begin{array}{l}{e^x} = \frac{{{x^0}}}{{0!}} + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} \\{e^x} = 1 + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} \end{array}\)

Subtract 1 from both sides

\({e^x} - 1 = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} \)

Divide both sides by x

\(\frac{{{e^x} - 1}}{x} = \sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} \)

Step 2: Integrate sides w.r.t x

\(\int {\frac{{{e^x} - 1}}{x}dx} = \sum\limits_{n = 1}^\infty {\int {\frac{{{x^{n - 1}}}}{{n!}}dx} } \)

In right side, we will use power rule for integration

\(\int {\frac{{{e^x} - 1}}{x}dx} = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n \cdot n!}}} + C\)