Here we have to find the value of ∫(x2+2)(x2+3)x2dx Let x2=t⇒(x2+2)(x2+3)x2=(t+2)(t+3)t Let (t+2)(t+3)t=t+2A+t+3B ⇒ t = A (t + 3) + B (t + 2) By putting t = - 2 on both the sides of (1) we get A = - 2 By putting t = - 3 on both the sides of (1) we get B = 3 (t+2)(t+3)t=t+2−2+t+33 By substituting x2=t in the above equation we get ⇒(x2+2)(x2+3)x2=x2+2−2+x2+33⇒∫(x2+2)(x2+3)x2dx=−2∫x2+2dx+3∫x2+3dx As we know that, ∫x2+a2dx=a1tan−1(ax)+C where C is a constant ⇒∫(x2+2)(x2+3)x2dx=2−2tan−1(2x)+33tan−1(3x)+C=−2tan−1(2x)+3tan−1(3x)+C Where C is a constant