Question Number 11429 by @ANTARES_VY last updated on 25/Mar/17

∫_0 ^3 ((2x+3)/(2x+1))dx=𝛂+ln7.  𝛂=?  please......

Answered by FilupS last updated on 26/Mar/17

((2x+3)/(2x+1))=((2x+1+2)/(2x+1))=1+(2/(2x+1))  ∫_0 ^( 3) ((2x+3)/(2x+1))dx=∫_0 ^( 3) 1+(2/(2x+1))dx  =∫_0 ^( 3) 1dx+∫_0 ^( 3) (2/(2x+1))dx  =[x]_0 ^3 +[ln(2x+1)]_0 ^3   =(3−0)+(ln(7)−ln(1))  =3+ln(7)     ∴α=3     red text is corrected

Commented by@ANTARES_VY last updated on 26/Mar/17

𝛂=2?...

Commented byFilupS last updated on 26/Mar/17

yes

Commented by@ANTARES_VY last updated on 26/Mar/17

such a wrong  answer....  no  𝛂=2?????

Commented byFilupS last updated on 26/Mar/17

∫_0 ^( 3) ((2x+3)/(2x+1))dx=α+ln(7)  as per my working:  ∫_0 ^( 3) ((2x+3)/(2x+1))dx=2+ln(7)  ∴α+ln(7)=2+ln(7)  α=2

Commented byajfour last updated on 26/Mar/17

come on       3+ln 7   it is.  ∫_0 ^3  ((2x+3)/(2x+1))dx = ∫_0 ^3 dx +∫_0 ^3  ((2dx)/(2x+1))

Commented byFilupS last updated on 26/Mar/17

I see my mistake.  I made a bad typo. Sorry