Question 1195861
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This is a calculus question. Could you explain question 18 on 
https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16600FE-F2016.pdf ? 
I know the answer is D but don't know how to get there. 
Please attach an image of your work as a reply.
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<pre>
Every/any/each Calculus student knows (or must know - this knowledge is a pre-requisite) that

    ln(1+y) = {{{y}}} - {{{y^2/2}}} + {{{y^3/3}}} - {{{y^4/4}}} + . . .     (1)

for all real values y, |y| < 1.  It is called  " the Maclaurin series of  f(y) = ln(1+y) ".



Replace (substitute) here y = -x^2.  You will get

    ln(1-x^2) = {{{-x^2}}} - {{{x^4/2}}} - {{{x^6/3}}} - {{{x^8/4}}} - . . .     (2)



Right side is your series with the sign "-, minus".



So, multiply both sides of identity (2) by (-1), and you will get the identity (D), which you need

    {{{-ln(1-x^2)}}} = {{{x^2}}} + {{{x^4/2}}} + {{{x^6/3}}} + {{{x^8/4}}} + . . .     (3)
</pre>

Solved and explained.