User:Cp wind55

$$ \int x^2 \ln x \ dx $$

$$ dv = x^2 \ dx \rightarrow v = \int x^2 \ dx = {x^3 \over 3} $$

$$ u = \ln x \rightarrow du = {1 \over x} \ dx $$

$$ \int u \ dv = uv - \int v \ du $$

$$ \int x^2 \ln x \ dx = {x^3 \over 3} \ln x - \int ({x^3 \over 3}) ({1 \over x}) \ dx $$

$$ = {x^3 \over 3} \ln x - {1 \over 3} \int x^2 \ dx $$

$$ = {x^3 \over 3} \ln x - {x^3 \over 9} + \ C $$

Evaluate $$ \int x^2 \ sin 4x \ dx $$

$$ \ sin 4x $$

$$ {-1 \over 4} \ cos 4x $$

$$ {-1 \over 16} \ sin 4x $$

$$ {1 \over 64} \ cos 4x $$

$$ \int x^2 \ sin 4x \ dx = {-1 \over 4} x^2 \ cos 4x + {1 \over 8} x \ sin 4x + {1 \over 32} \ cos 4x + \ C $$