Question

How do you find `\int 4x ^ { 2} e ^ { 2x } d x`?

Answer 1

`(2x^2-2x+1)e^(2x)+C.`

Explanation:

Let `int4x^2e^(2x)dx=4intx^2e^(2x)dx=4I, where, I=intx^2e^(2x)dx.`

We use the Rule of Integration by Parts (IBP) in the following

Form :

IBP : `intuvdx=uintvdx-int{(du)/dx*intvdx}dx.`

We take, `u=x^2, and, v=e^(2x).`

`:. (du)/dx=2x, and, intvdx=1/2e^(2x).`

`:. I=x^2(1/2e^(2x))-int{(2x)(1/2e^(2x))}dx,`

`=1/2*x^2e^(2x)-intxe^(2x)dx=1/2*x^2e^(2x)-J, where,`

`J=intxe^(2x)dx,` and, reapplying IBP with, `u=x, v=e^(2x),`

`J=x(1/2e^(2x))-int{(1)(1/2e^(2x))}dx,`

`=1/2*xe^(2x)-1/2inte^(2x)dx,`

`=1/2*xe^(2x)-1/2(1/2e^(2x)).`

`rArr J=1/2*xe^(2x)-1/4e^(2x).`

`:. I=1/2*x^2e^(2x)-1/2*xe^(2x)+1/4e^(2x).`

`:. int4x^2e^(2x)dx=4{1/2*x^2e^(2x)-1/2*xe^(2x)+1/4e^(2x)},`

`=2x^2e^(2x)-2xe^(2x)+e^(2x).`

`rArr int4x^2e^(2x)dx=(2x^2-2x+1)e^(2x)+C.`

Enjoy Maths.!

Answer 2

`int \ 4x^2e^(2x) \ dx = (2x^2 -2x+ 1)e^(2x) + c`

Explanation:

Another approach if you are keen to avoid Integration By Parts is to work backward from the derivative of a possible solution.

Knowing that IBP will reduce an integral involving `x^2e^(2x)` to an integral involving `xe^(2x)` and a further application will reduce an integral involving `xe^(2x)` to one involving `e^(2x)` then we would expect a solution of the form:

`int \ 4x^2e^(2x) \ dx = (Ax^2 + Bx+ C)e^(2x) + c`

Differentiating wrt `x` whilst applying the product rule, we get:

`4x^2e^(2x) = (Ax^2 + Bx+ C)2e^(2x) + (2Ax + B)e^(2x)`
`" " = (2Ax^2 + (2B+2A)x+ B+2C)e^(2x)`

Equating coefficients, we get:

`x^2 : \ 4 = 2A \ \ \ \ \ \ \ \ \ \=> A=2`
`x^1 : \ 0 = 2B+2A => B=-2`
`x^0 : \ 0 = B+2C \ \ => C=1`

Thus the solution is:

`int \ 4x^2e^(2x) \ dx = (2x^2 -2x+ 1)e^(2x) + c`