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

1 Answer
Apr 18, 2018

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))