Question

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

How do you evaluate this indefinite integral as a power series?

`int ( 8(e^x -1))/(7x) dx`

Answer 1

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

Explanation:

Rewrite the integral a bit:

`int(8e^x-8)/(7x)dx=int8/7x^-1e^x-8/7x^-1 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^-1x^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 `8/7x^-1` vanish:

The `0th` 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))`