What is the integral of (9x^2-4x)e^{2x} ?

The integral of (9x^2-4x)e^{2x} is 1/4 e^{2x}(18x^2-26x+13)+C

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integral of (9x^2-4x)e^{2x}

\int (9 x^{2} - 4 x) e^{2 x} d x

\frac{1}{4} e^{2 x} (18 x^{2} - 26 x + 13) + C

Solution steps

\int (9 x^{2} - 4 x) e^{2 x} d x

Apply u-substitution

: \int \frac{9 ln ^{2} ( u ) - 8 ln ( u )}{8} d u

= \int \frac{9 ln ^{2} ( u ) - 8 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 9 ln^{2} (u) - 8 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 9 ln^{2} (u) d u - \int 8 ln (u) d u)

\int 9 ln^{2} (u) d u = 9 (u ln^{2} (u) - 2 (u ln (u) - u))

\int 8 ln (u) d u = 8 (u ln (u) - u)

= \frac{1}{8} (9 (u ln^{2} (u) - 2 (u ln (u) - u)) - 8 (u ln (u) - u))

Substitute back

u = e^{2 x}

= \frac{1}{8} (9 (e^{2 x} ln^{2} (e^{2 x}) - 2 (e^{2 x} ln (e^{2 x}) - e^{2 x})) - 8 (e^{2 x} ln (e^{2 x}) - e^{2 x}))

Simplify

\frac{1}{8} (9 (e^{2 x} ln^{2} (e^{2 x}) - 2 (e^{2 x} ln (e^{2 x}) - e^{2 x})) - 8 (e^{2 x} ln (e^{2 x}) - e^{2 x})) : \frac{1}{4} e^{2 x} (18 x^{2} - 26 x + 13)

= \frac{1}{4} e^{2 x} (18 x^{2} - 26 x + 13)

Add a constant to the solution

= \frac{1}{4} e^{2 x} (18 x^{2} - 26 x + 13) + C

\int e^{- 4 x} x d x

\int \frac{2 x + 3}{2 x + 1} d x

\frac{\partial}{\partial y} (e^{x + y})

x \to 2 lim (9 - x^{2})

n = 1 \sum \infty \frac{7}{n ^{2} + 4}

What is the integral of (9x^2-4x)e^{2x} ?
The integral of (9x^2-4x)e^{2x} is 1/4 e^{2x}(18x^2-26x+13)+C