ज्ञात करें `int x^2 log (1+x) dx`

Explanation

`I=int x^2 log ( 1+x )dx = log ( 1+x ) x^3/3 - int 1/(1+x)x^3/3 ""dx `
`x^3/3 "log (1+x)"-1/3 int x^3/(1+x)dx`
अब, `int x^3/(1+x)dx = int (x^3 +1-1)/(x+1)dx=int (x^3+1)/(x+1)dx- int (dx)/(x+1)`
`=int (x^2-x+1)dx - int (dx)/(x+1)=x^3/(x+1)=x^3/3-x^2/2+x - log (x+1) +c" "`[यहां `(x-1)` क्योँकि `log (x+1)` परिभाषित है ]
`:." "` (1) से, `I =x^3/3"log (1+x)"-1/3[x^3/3-x^2/2+x-log ( 1+x)]+c`