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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

= \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 r - 1

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

Simplify

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

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

Add a constant to the solution

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

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