# Evaluate int ( 8(e^x -1))/(7x) dx as a power series.?

## How do you evaluate this indefinite integral as a power series? $\int \frac{8 \left({e}^{x} - 1\right)}{7 x} \mathrm{dx}$

Apr 18, 2018

C+8/7sum_(n=1)^oox^n/(n!(n))

#### Explanation:

Rewrite the integral a bit:

$\int \frac{8 {e}^{x} - 8}{7 x} \mathrm{dx} = \int \frac{8}{7} {x}^{-} 1 {e}^{x} - \frac{8}{7} {x}^{-} 1 \mathrm{dx}$

Now, recall the power series expansion for ${e}^{x}$ about $a = 0 :$

e^x=sum_(n=0)^oox^n/(n!)

Thus, we rewrite as

int8/7x^-1sum_(n=0)^oox^n/(n!)-8/7x^-1 dx

We may multiply in the ${x}^{-} 1$ into the series, as ${x}^{-} 1 {x}^{n} = {x}^{n - 1}$

int8/7sum_(n=0)^oox^(n-1)/(n!)-8/7x^-1 dx

Let's evaluate the series at $n = 0$ to see if we can make $\frac{8}{7} {x}^{-} 1$ vanish:

The $0 t h$ term of the series is 8/7x^-1/(0!)=8/7x^-1.

So, we have

int(cancel(8/7x^-1)+8/7sum_(n=1)^oox^(n-1)/(n!)-cancel(8/7x^-1))dx

int8/7sum_(n=1)^oox^(n-1)/(n!)dx=sum_(n=1)^oo8/7intx^(n-1)/(n!)dx

Term-by-term integration yields

C+8/7sum_(n=1)^oox^n/(n!(n))