Question

integration of (x^3)e^3x​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int x^{3} e^{3x} dx$

As we know that,

In this type of question we can apply integration by parts, we get.

(1) = If u and v are two functions of x then,

∫(u v)dx = u ∫v dx - ∫[(du/dx).∫v dx ]dx.

From the first order of letters.

I = Inverse trigonometric functions.

L = Logarithmic functions.

A = Algebraic functions.

T = Trigonometric functions.

E = Exponential functions.

This is known as ILATE.

Therefore, arrange the functions in the order according to letters of this word and then integrate by parts.

Apply this rule in the integration, we get.

x³ = considered as first functions = Algebraic functions.

e³ˣ = considered as second functions = Exponential functions.

$\sf \implies x^{3} \displaystyle \int ex^{3x} dx \ - \int \bigg[\dfrac{d(x^{3}) }{dx} . \int e^{3x}dx \bigg] dx$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \int \bigg[3x^{2} . \dfrac{e^{3x} }{3} \bigg] dx$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \int \bigg[ x^{2} .{e^{3x} }{} \bigg]dx$

Apply integration by parts on ∫x².e³ˣ dx, we get.

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \bigg[x^{2} .\int e^{3x} \ - \int \bigg(\dfrac{d(x^{2} )}{dx} . \int e^{3x} dx \bigg)dx \bigg].$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \bigg[x^{2} .\dfrac{e^{3x} }{3} \ - \int \bigg(2x .\dfrac{e^{3x} }{3} \bigg)dx \bigg].$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \bigg[\dfrac{x^{2} e^{3x} }{3} \ - \dfrac{2}{3} \int (x. e^{3x})dx \bigg].$

As we know that,

Again apply integration by parts : ∫x.e³ˣ dx, we get.

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \dfrac{x^{2} e^{3x} }{3} + \dfrac{2}{3} \bigg[x \int e^{3x} \ - \int \bigg[\bigg(\dfrac{dx}{dx} \bigg) . \int e^{3x} dx \bigg] \bigg].dx$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \dfrac{x^{2} e^{3x} }{3} + \dfrac{2}{3} \bigg[ x. \dfrac{e^{3x} }{3} \ - \int \dfrac{e^{3x} }{3} dx \bigg]$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \dfrac{x^{2} e^{3x} }{3} + \dfrac{2}{3} \bigg[ \dfrac{xe^{3x} }{3} \ - \dfrac{e^{3x} }{9} \bigg] + c.$

$\sf \implies x^{3} \displaystyle \ \dfrac{e^{3x} }{3} \ - \dfrac{x^{2} e^{3x} }{3} + \dfrac{2xe^{3x} }{9} \ - \dfrac{2e^{3x} }{27} + c.$

$\sf \implies e^{3x} \bigg(\dfrac{x^{3} }{3} \ - \dfrac{x^{2} }{3} \ + \dfrac{2x}{9} \ - \dfrac{2}{27} \bigg) + c.$

$\sf \implies \displaystyle \int x^{3} e^{3x} dx = e^{3x} \bigg(\dfrac{x^{3} }{3} \ - \dfrac{x^{2} }{3} \ + \dfrac{2x}{9} \ - \dfrac{2}{27} \bigg) + c.$

                                                                                                                         

MORE INFORMATION.

Standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, ( k ∈ R ).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.

Answer 2

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, ( k ∈ R ).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.