What is the integral of (2w^2-w)e^{7w-1} ?

The integral of (2w^2-w)e^{7w-1} is 1/343 (98e^{7w-1}w^2-77e^{7w-1}w+11e^{7w-1})+C

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integral of (2w^2-w)e^{7w-1}

\int (2 w^{2} - w) e^{7 w - 1} d w

\frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1}) + C

Solution steps

\int (2 w^{2} - w) e^{7 w - 1} d w

Apply u-substitution

: \int \frac{e ^{u} ( 2 u ^{2} - 3 u - 5 )}{343} d u

= \int \frac{e ^{u} ( 2 u ^{2} - 3 u - 5 )}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int e^{u} (2 u^{2} - 3 u - 5) d u

Expand

e^{u} (2 u^{2} - 3 u - 5) : 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{u}

= \frac{1}{343} \cdot \int 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{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}{343} (\int 2 e^{u} u^{2} d u - \int 3 e^{u} u d u - \int 5 e^{u} d u)

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

\int 3 e^{u} u d u = 3 (e^{u} u - e^{u})

\int 5 e^{u} d u = 5 e^{u}

= \frac{1}{343} (2 (u^{2} e^{u} - 2 (e^{u} u - e^{u})) - 3 (e^{u} u - e^{u}) - 5 e^{u})

Substitute back

u = 7 w - 1

= \frac{1}{343} (2 ((7 w - 1)^{2} e^{7 w - 1} - 2 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1})) - 3 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1}) - 5 e^{7 w - 1})

Simplify

\frac{1}{343} (2 ((7 w - 1)^{2} e^{7 w - 1} - 2 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1})) - 3 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1}) - 5 e^{7 w - 1}) : \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1})

= \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1})

Add a constant to the solution

= \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1}) + C

\int \frac{4 - x ^{2}}{2 + x} d x

\frac{d y}{d t} = 3 y

\int \frac{4}{x} - 7 e^{x} d x

\int 8 t e^{7 t} d t

y = x^{3 - 5 x^{2}} + 7 x + 3

What is the integral of (2w^2-w)e^{7w-1} ?
The integral of (2w^2-w)e^{7w-1} is 1/343 (98e^{7w-1}w^2-77e^{7w-1}w+11e^{7w-1})+C