Integration of Logs (1 Viewer)

[poopoohead]

Hey, im having trouble integrating this:

The integral from 2 to 1 of:

(2x³ - x² + 5x + 3)/x




The answer is: 8 and 1/6 + 3ln2


can someone maybe give me some help- thanks <o></o>


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[poopoohead]

don't know what all those smiley faces are doing there

 

[who_loves_maths]

Hey,

Divide the 'x' throughout each term of the numerator:

ie. (2x³ - x² + 5x + 3)/x = 2x³/x - x²/x + 5x/x + 3/x = 2x² - x + 5 + 3/x

And so:

I = {2 -> 1}∫(2x³ - x² + 5x + 3)/x dx = ∫2x² dx - ∫x dx + ∫5 dx + ∫3/x dx
= (2/3)[x³]{2 ->1} - (1/2)[x²]{2 ->1} + 5[x]{2 ->1} + 3[ln(x)]{2 ->1}

= 49/6 + 3ln(2)

as required.

 

[poopoohead]

thank you