ज्ञात करें `int (2-3x)/(sqrt(1+3x))dx`

Explanation

यहां `P(x)=2-3x` तथा `sqrt(1+3x),(ax+b)^n` के रूप का है |
माना की `z=1-3x` तो `dz=3dx" ":." "dx=(dz)/3`
पुनः `z=1+3x" ":." "x=(z-1)/3`
अब, `int(2-3x)/(sqrt(1+3x))dx=int(2-3((z-1)/x))/sqrtz(dz)/3`
`=1/3int(2-z+1)/sqrtzdx=1/3int(3/sqrt3-z/sqrtz)dz`
`=1/3int(3z^(1/2)-sqrt2)dz`
`=1/3[(3z^(1/2))/(1/2)-z^(3/2)/(3/2)+c=1/3(6sqrtz-2/3z^(3/2))]+c`
`=(2sqrt3)/3(3-1/3z)+c=2/3sqrtz((9-z))/3+c`
`=2/3sqrt(1+3x)[(9-(1+3x))/3]+c`
`=2/3sqrt(1+3x)((8-3x)/3)+c=2/3(8-3x)sqrt(1+3x)+c`