Integration using partial fractions

[@vaibhavimal]

very helpul video

[@adityavyas2156]

Thanks bro.Got IIT because of your sexy techniques

[@manjimaspramod6502]

Thank you so much Sir🤗

[@deepanshugupta9259]

Easy peasy lemon squeezy🥳🥳

[@manjimaspramod6502]

True

[@hudagal6471]

This lightened up my night even before I watched the video 😂😂😂😂😂

[@adityaprakashgiri6462]

Thank you very much sir

[@kasandabinwell4376]

Thanks for your work

[@kevenkoshie540]

good one

[@LeviousBanda]

thank you sir

[@matengekanini926]

Hizo numbers za denomenator Ulipata aje

[@dragobeastyt3142]

It's an wrong answer sir

[@sauvejean-luc3814]

It's not 1/2 before ln,
It's 2.

[@InshalHassan-r4s]

@@sauvejean-luc3814 That's what I am thinking

[@gracecook7661]

@@InshalHassan-r4s The 1/2 is correct. To integrate 2/(4x-1), you use substitution for the 4x-1 (he just didn't show the steps). u = 4x - 1, du = 4dx. S 2/(4x-1) dx = 2 S 1/(4x-1) *1/4 * 4 dx = 2 *1/4 S 1/u du = 1/2 S 1/u du = 1/2 ln |u| = 1/2 ln |4x-1|

[@sumitkumarghosh4867]

@@sauvejean-luc3814 Thanks man! I also got confused

[@ArjunDOnlyHero]

@@gracecook7661 or simply, since integration of 1/(ax+b) = (1/a) × ln|ax+b|, integration of 2/(4x-1) = 2 [(1/4) × ln|4x-1|.
Thus, it becomes ½ ln|4x-1|.

[@robiesther8910]

You are too fast mr