f(x)+c Simplify the above equation as follows x3⋅‌
e2x
2
−∫3x2⋅‌
e2x
2
dx=‌
e2x
8
f(x)+c ‌
1
2
x3e2x−‌
3
2
[x2‌
e2x
2
−∫2x⋅‌
e2x
2
dx]=‌
e2x
8
f(x)+c ‌
1
2
x3e2x−‌
3
4
x2e2x+‌
3
2
‌∫x⋅e2xdx=‌
e2x
8
f(x)+c ‌
1
2
x3e2x−‌
3
4
x2e2x+‌
3
2
[x‌
e2x
2
−∫1⋅‌
e2x
2
dx]=‌
e2x
8
f(x)+c Simplify further, ‌
e2x
8
[4x3−6x2+6x−3]+c1=‌
e2x
8
f(x)+c Therefore, on comparison f(x)‌‌=4x3−6x2+6x−3 1‌‌=4x3−6x2+6x−3 4x3−6x2+6x−4‌‌=0 2(x3−1)−3x(x−1)‌‌=0Solve further 2(x−1)(x2+x+1)−3x(x−1)‌‌=0 (x−1)(2x2+2x+2−3x)‌‌=0 (x−1)(2x2−x+2)‌‌=0 x=1 is a real root. 1 The sum of non-real complex roots from the above equation is 2