∫x2+x+53x+2dx
⇒∫x23x+2dx+∫x3x+2dx+∫53x+2dx
⇒∫(x3−29+49(3x+2))dx
I1=∫x3dx+∫−29dx+∫49(3x+2)dx
⇒x26−29x+49∫13x+2dx
49.3ln|3x+2|
I2:∫x3x+2dx
∫19(U−2U)dv
19U−29ln|u|⇒19((3x+2)−2ln/3x+2)
I3⇒∫53x+2dx
=53ln|3x+2|
I=x26−2x9+427ln|3x+2|+19(3x+2−2n(3x+2)+53ln|3x+2|
⇒x26+19x+4327ln|3x+2|+29