What is the integral of 6x^2e^{6x} ?

The integral of 6x^2e^{6x} is e^{6x}x^2-1/3 e^{6x}x+1/18 e^{6x}+C

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integral of 6x^2e^{6x}

\int 6 x^{2} e^{6 x} d x

e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x} + C

Solution steps

\int 6 x^{2} e^{6 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x^{2} e^{6 x} d x

Apply Integration By Parts

= 6 (\frac{1}{6} e^{6 x} x^{2} - \int \frac{1}{3} e^{6 x} x d x)

\int \frac{1}{3} e^{6 x} x d x = \frac{1}{108} (6 e^{6 x} x - e^{6 x})

= 6 (\frac{1}{6} e^{6 x} x^{2} - \frac{1}{108} (6 e^{6 x} x - e^{6 x}))

Simplify

6 (\frac{1}{6} e^{6 x} x^{2} - \frac{1}{108} (6 e^{6 x} x - e^{6 x})) : e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x}

= e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x}

Add a constant to the solution

= e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x} + C

a re a y = x^{3} + x^{2} - 6 x, y = 6 x

l a pl a ce t r an s f or m e^{(3 t)} (4 cos (3 t) - \frac{10}{3} sin (3 t))

x \to - 1 + lim (\frac{x}{∣ x + 1 ∣})

so l v e f or y, y = x e^{y}

\int_{- 3}^{- 2} (2 x (3 - x^{2})^{3}) d x

What is the integral of 6x^2e^{6x} ?
The integral of 6x^2e^{6x} is e^{6x}x^2-1/3 e^{6x}x+1/18 e^{6x}+C