Question

Integrate £x² e3x dx​

Answer 1

Answer:

Refer the attachement for better understanding!

➡️Use integration by parts method.

Be Brainly!

Answer 2

EXPLANATION.

Integration.

⇒ ∫x².e³ˣ dx.

As we know that,

In this question we can apply integration by-parts.

If u and v are two functions of x then,

$\implies \displaystyle \int (u.v)dx = u .\int v dx \ - \int \bigg[ \dfrac{du}{dx} . \int v dx \bigg] dx$

From the first letter.

I = Inverse trigonometric function.

L = Logarithmic functions.

A = Algebraic functions.

T = Trigonometric functions.

E = Exponential functions.

We get a word = ILATE.

In this equation,

⇒ x² = first functions.

⇒ e³ˣ = second functions.

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

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

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

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

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

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

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

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

$\implies \displaystyle e^{3x} \bigg[ \dfrac{x^{2} }{3} - \dfrac{2.x}{9} + \dfrac{2}{27} \bigg] + C.$

$\implies \displaystyle \int x^{2} .e^{3x} dx \ = e^{3x} \bigg[ \dfrac{x^{2} }{3} - \dfrac{2.x}{9} + \dfrac{2}{27} \bigg] + C.$