Question

Evaluate the indefinite integral as an infinite series.

A)

Evaluate the indefinite integral as an infinite series. 5 ex - 1/8x dx

Answer 1

Need to find \(5 \int \frac{e^{x}-1}{8 x} d x\).

Recoll ect that,

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

Therefore,

$$ \begin{aligned} \frac{e^{x}-1}{x} &=\frac{\left(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots+\frac{x^{n}}{n !}+\ldots\right)-1}{x} \\ &=\frac{\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots+\frac{x^{n}}{n !}+\ldots}{x} \\ &=\frac{1}{1 !}+\frac{x}{2 !}+\frac{x^{2}}{3 !}+\ldots+\frac{x^{n-1}}{n !}+\ldots \end{aligned} $$

Now integrating to get

$$ \begin{aligned} \int \frac{e^{x}-1}{x} d x &=\int\left(\frac{1}{1 !}+\frac{x}{2 !}+\frac{x^{2}}{3 !}+\ldots+\frac{x^{n-1}}{n !}+\ldots\right) d x \\ &=\frac{x}{1 !}+\frac{x^{2}}{2 \cdot 2 !}+\frac{x^{3}}{3 \cdot 3 !}+\ldots+\frac{x^{n}}{n \cdot n !}+\ldots+C \\ &=\sum_{n=1}^{\infty} \frac{x^{n}}{n \cdot n !}+C \\ \text { Thus, } 5 \int \frac{e^{x}-1}{8 x} d x &=\frac{5}{8} \int \frac{e^{x}-1}{x} d x \\ &=\sum_{n=1}^{\infty} \frac{5}{8} \cdot \frac{x^{n}}{n \cdot n !}+C \end{aligned} $$