Evaluate the indefinite integral as an infinite series. A) Evaluate the indefinite integral as an infinite…
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Question “Evaluate the indefinite integral as an infinite series. A) Evaluate the indefinite integral as an infinite…”
Evaluate the indefinite integral as an infinite series.
A)
Answer
You need to locate
Recoll ect.
Therefore,
Now integrate to get
5 \int \frac{e^{x}-1}{8 x} d x
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 !}+. .
\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}
\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}
Conclusion
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