The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

core 3 integration

upper limit: (e/2) and lower limit: 1.


integrate (ln2x)/x^4 dx
Reply 1
Avatar for a.myhall
a.myhall
OP
6
Original post by a.myhall
upper limit: (e/2) and lower limit: 1.


integrate (ln2x)/x^4 dx


The answer is (3ln2 + 1 -32e^-3)/9 by the way.
Original post by a.myhall
upper limit: (e/2) and lower limit: 1.


integrate (ln2x)/x^4 dx


Integrate by parts, 'u' should always be the ln term.
Let u=ln(2x)
Let dv/dx=x^-4
Reply 4
Avatar for a.myhall
a.myhall
OP
6
yep I've done that and tried to integrate, getting (-(ln2x/3x^3) -1/x^3)... not sure if this is right and I am unable to put the limits in and get the correct answer? Can anyone write out their working for me please?
Original post by a.myhall
yep I've done that and tried to integrate, getting (-(ln2x/3x^3) -1/x^3)... not sure if this is right and I am unable to put the limits in and get the correct answer? Can anyone write out their working for me please?


∫1e/2x−4ln⁡(2x).dx=[−13x−3ln⁡(2x)]1e/2−∫1e/2−x−33x.dx\displaystyle \int_1^{e/2} x^{-4}\ln(2x) .dx = \left[ -\frac{1}{3}x^{-3}\ln(2x) \right]_1^{e/2} - \int_1^{e/2} -\frac{x^{-3}}{3x}.dx

=[−13x−3ln⁡(2x)]1e/2−[19x−3]1e/2\displaystyle =\left[ -\frac{1}{3}x^{-3}\ln(2x) \right]_1^{e/2} - \left[ \frac{1}{9}x^{-3} \right]_1^{e/2}
(edited 7 years ago)
Original post by a.myhall
yep I've done that and tried to integrate, getting (-(ln2x/3x^3) -1/x^3)... not sure if this is right and I am unable to put the limits in and get the correct answer? Can anyone write out their working for me please?


http://prnt.sc/eucj43

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you