Question Number 100657 by Cheyboy last updated on 28/Jun/20

∫  ((3x−1)/(x^2 +9)) dx

Commented bybobhans last updated on 28/Jun/20

set x = 3tan z ⇒ dx = 3sec ^2 dz  I=∫ ((2x+x−1)/(x^2 +9)) dx =∫ ((d(x^2 +9))/(x^2 +9))+∫ (((3tan z−1)3sec ^2 z dz)/(9sec ^2 z))  =ln(x^2 +9)−(1/3)∫(3((d(cos z))/(cos z)))−∫(1/3)dz  =ln(x^2 +9)−ln∣cos z∣−(1/3)z + c   now substitute tan z = (x/3)

Commented byDwaipayan Shikari last updated on 28/Jun/20

(3/2)∫((2x)/(x^2 +9))dx−∫(1/(x^2 +9))dx  (3/2)log(x^2 +9)−(1/3)tan^(−1) (x/3)+constant

Commented byCheyboy last updated on 28/Jun/20

Thank all for the help

Answered by mathmax by abdo last updated on 28/Jun/20

I =∫ ((3x−1)/(x^2  +9))dx ⇒ I =(3/2)∫ ((2x)/(x^2  +9))dx −∫ (dx/(x^2  +9))(→x =3t)  =(3/2)ln(x^2  +9)−∫ ((3dt)/(9(1+t^2 ))) =(3/2)ln(x^2  +9)−(1/3) arctant +C  =(3/2)ln(x^2  +9)−(1/3) arctan((x/3)) +C