Write the integrand as a power series. Use this to evaluate the indefinite integral as a series. 2*integral[((e^x)-1)/(7x)] dx. Please help! Ty

Question

anonymous · Answer

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anonymous · Answer

Those are the options

anonymous · Answer

This is how I tried to do it, but I think my manipulation is wrong.

reemii · Answer

The series of the exponential starts at 0: $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$.
So $(e^x-1)/7x$ is $\frac 17 \sum_{k=1}^\infty \frac{x^{k-1}}{k!}$.

reemii · Answer

oh, you didn't get it wrong for the "start at zero" remark, my bad.
I think the answer is C.

anonymous · Answer

Thank you! Do you might telling me what I did wrong. Since there was a -1 on the left side of the equation, i added a -1 on the right… was this wrong?

reemii · Answer

This was wrong indeed. The effect of the "-1" is to change the n=0 to n=1 in the sum.

anonymous · Answer

oh yeah! I get it now! :) Thank you!