What is the integral of (9x^2-8x)e^{2x} ?

The integral of (9x^2-8x)e^{2x} is 1/4 e^{2x}(18x^2-34x+17)+C

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integral of (9x^2-8x)e^{2x}

\int (9 x^{2} - 8 x) e^{2 x} d x

\frac{1}{4} e^{2 x} (18 x^{2} - 34 x + 17) + C

Solution steps

\int (9 x^{2} - 8 x) e^{2 x} d x

Apply u-substitution

: \int \frac{9 ln ^{2} ( u ) - 16 ln ( u )}{8} d u

= \int \frac{9 ln ^{2} ( u ) - 16 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) - 16 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 16 ln (u) d u)

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

\int 16 ln (u) d u = 16 (u ln (u) - u)

= \frac{1}{8} (9 (u ln^{2} (u) - 2 (u ln (u) - u)) - 16 (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})) - 16 (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})) - 16 (e^{2 x} ln (e^{2 x}) - e^{2 x})) : \frac{1}{4} e^{2 x} (18 x^{2} - 34 x + 17)

= \frac{1}{4} e^{2 x} (18 x^{2} - 34 x + 17)

Add a constant to the solution

= \frac{1}{4} e^{2 x} (18 x^{2} - 34 x + 17) + C

y^{^{''}} - 6 y + 9 y = 0

\frac{d y}{d x} = \frac{( 2 x ^{2} + 3 ) ( 3 x + 2 )}{7 x - 3}

(x + 1) \frac{d y}{d x} = x + 6

\frac{d}{d x} (co tan (2 x))

\int \frac{1}{7 + 9 x ^{2}} d x

What is the integral of (9x^2-8x)e^{2x} ?
The integral of (9x^2-8x)e^{2x} is 1/4 e^{2x}(18x^2-34x+17)+C