What is the integral of (7x^2-12x)e^{2x} ?

The integral of (7x^2-12x)e^{2x} is 1/4 e^{2x}(14x^2-38x+19)+C

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integral of (7x^2-12x)e^{2x}

\int (7 x^{2} - 12 x) e^{2 x} d x

\frac{1}{4} e^{2 x} (14 x^{2} - 38 x + 19) + C

Solution steps

\int (7 x^{2} - 12 x) e^{2 x} d x

Apply u-substitution

: \int \frac{7 ln ^{2} ( u ) - 24 ln ( u )}{8} d u

= \int \frac{7 ln ^{2} ( u ) - 24 ln ( u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 7 ln^{2} (u) - 24 ln (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int 7 ln^{2} (u) d u - \int 24 ln (u) d u)

\int 7 ln^{2} (u) d u = 7 (u ln^{2} (u) - 2 (u ln (u) - u))

\int 24 ln (u) d u = 24 (u ln (u) - u)

= \frac{1}{8} (7 (u ln^{2} (u) - 2 (u ln (u) - u)) - 24 (u ln (u) - u))

Substitute back

u = e^{2 x}

= \frac{1}{8} (7 (e^{2 x} ln^{2} (e^{2 x}) - 2 (e^{2 x} ln (e^{2 x}) - e^{2 x})) - 24 (e^{2 x} ln (e^{2 x}) - e^{2 x}))

Simplify

\frac{1}{8} (7 (e^{2 x} ln^{2} (e^{2 x}) - 2 (e^{2 x} ln (e^{2 x}) - e^{2 x})) - 24 (e^{2 x} ln (e^{2 x}) - e^{2 x})) : \frac{1}{4} e^{2 x} (14 x^{2} - 38 x + 19)

= \frac{1}{4} e^{2 x} (14 x^{2} - 38 x + 19)

Add a constant to the solution

= \frac{1}{4} e^{2 x} (14 x^{2} - 38 x + 19) + C

x \to - 2 - lim (- 3 x^{2} + 2)

y^{^{'}} = \frac{1}{2} + \frac{1}{4} sin (t) - \frac{y}{50}

\int \frac{a}{b ^{3 θ}} d θ

x \to 0 lim (\frac{sin ( x ^{0} )}{x})

What is the integral of (7x^2-12x)e^{2x} ?
The integral of (7x^2-12x)e^{2x} is 1/4 e^{2x}(14x^2-38x+19)+C