Answer:

Step-by-step explanation:
<u>Fundamental Theorem of Calculus</u>

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.
Given indefinite integral:

Take <u>partial fractions</u> of the given fraction by writing out the fraction as an <u>identity</u>:

Calculate the values of A and B using substitution:


Substitute the found values of A and B:


If the terms are multiplied by constants, take them outside the integral:



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