Integrate 1/(5-2x) for 3≤x≤4

You must be careful with these sorts of questions as although 1/(5-2x) is equivalent to (5-2x)^-1, when you integrate you would add one to the power and divide by the new power. But if you were to add one to (5-2x)^-1 you would get zero. Therefore, when you are integrating a fraction with a linear expression as the denominator (meaning a denominator where the greatest power of x is 1), it integrates to the natural logarithm (ln) of the denominator, multiplied by the differential of the denominator.

So in this example, 1/(5-2x) would integrate to

[ln(5-2x)/(-2)] (as 5-2x differentiates to -2) for 3≤x≤4

Then you would sub in the limits of x and subtract as usual:

ln(5-2(4))/(-2) - ln(5-2(3))/(-2)

= -1/2ln(5-8) - -1/2ln(5-6) 

Remeber that you cannot take the ln of a negative number, so it is best to write it as: 

= -1/2ln|-3| - -1/2ln|-1|

= -1/2ln(3) - -1/2ln(1)

ln(1)=0 so our answer is



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