How do you find the antiderivative of $x^{2} e^{2 x}$ $x^2 e^(2x)$ ?

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Answer 1 · 2018-06-03T13:41:50

$\frac{1}{2} e^{2 x} (x^{2} - x + \frac{1}{2}) + C$ $1/2e^(2x)(x^2 - x + 1/2) + C$

Well, This can't be done simply.

We have to Perform Integration By Parts.

The Rule States That,

$int uvdx = uintvdx - int (u'intv)dx$

Where $u$ and $v$ are functions of $x$ and $u' = d/dx(u)$ .

Now, We have,

$intx^2e^(2x)dx$

$= x^2inte^2x - int(d/dx(x^2)inte^(2x))dx$

$= 1/2x^2e^(2x) - int(cancel(2)x xx 1/cancel(2)e^(2x))dx$

$= 1/2x^2e^(2x) - int xe^(2x) dx$ ......................(i)

We have to Integrate By Parts Once More.

So, From (i),

$1/2x^2e^(2x) - x inte^(2x) + int(d/dx(x)inte^(2x))dx$

$= 1/2x^2e^(2x) - 1/2xe^(2x) + 1/2inte^(2x)dx$

$= 1/2x^2e^(2x) - 1/2xe^(2x) + 1/4e^(2x) + C$ [Yeah, Don't EVER Forget this guy here $rarr$ $C$ ]

$= 1/2e^(2x)(x^2 - x + 1/2) + C$

Hope this helps.

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