∫x3+3x+4√xdx
⇒∫(x3−1/2+3x1−1/2+4x−1/2)dx
⇒∫x5/2+3x1/2+4x−1/2
⇒x7/27/2+x3/23/2+8x1/2+c
(II)
∫√x(3x2+2x+3)dx
⇒∫(3x5/2+2x3/2+3x1/2)dx
⇒3x7/27/2+2x5/25/2+3x3/23/2+c
(III)
∫(2x2−3sinx+5√x)dx
⇒2x33+3cosx+5x3/23/2+c
(IV)
∫sec2xcosec2xdx
=∫tan2xdx
⇒∫(sec2x+1)dx
⇒tanx−x+c