∫ x4 e2x dx =

Question

∫ x4 e2x dx =  (A) (e2 x/4) (2 x4 - 4x3 + 6 x2 - 6 x + 3 )+ C  (B) (e 2x/2) (2 x4 - 4 x3 + 6 x2 - 6 x + 3 )+ C  (C) (e2 x/8) (2 x4 + 4 x3 + 6x2 + 6 x + 3 )+ C  (D) (e2x/4) (2 x4 + 4 x3 + 6 x2 + 6 x + 3 )+ C

Tardigrade Admin · Accepted Answer

Let I=∫ x4 e2 x d x ⇒ I =x4 ∫ e2 x d x-∫ (d x4/d x) ⋅ ∫ e2 x ⋅ d x ⋅ d x+C =(x4 ⋅ e2 x/2)-(1/2) ∫ 4 x3 e2 x d x+C ⇒ (x4 e2 x/2)-2[(x3 e2 x/2)-∫ (3 x2 e2 x/2) d x]+C ⇒ (x4 e2 x/2)-x3 e2 x+3[(x2 e2 x/2)-(1/2) ∫ 2 x e2 x d x]+C ⇒ (x4 e2 x/2)-x3 e2 x+(3/2) x2 e2 x-3 [(x e2 x/2)-∫ (e2 x/2) d x]+C ⇒ (x4 e2 x/2)-x3 e2 x+(3/2) x2 e2 x-(3/2) x e2 x+(3 e2 x/4)+C ⇒ (e2 x/4)[2 x4-4 x3+6 x2-6 x+3]+C

Q. $\int x^{4} e^{2 x} d x =$

$\frac{e ^{2 x}}{4} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$

$\frac{e 2 x}{2} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$

$\frac{e ^{2 x}}{8} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C$

$\frac{e ^{2 x}}{4} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C$

Q. ∫x4e2xdx=

4e2x​(2x4−4x3+6x2−6x+3)+C

2e2x​(2x4−4x3+6x2−6x+3)+C

8e2x​(2x4+4x3+6x2+6x+3)+C

4e2x​(2x4+4x3+6x2+6x+3)+C

Q. $\int x^{4} e^{2 x} d x =$

$\frac{e ^{2 x}}{4} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$

$\frac{e 2 x}{2} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$

$\frac{e ^{2 x}}{8} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C$

$\frac{e ^{2 x}}{4} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C$