What is the integral of (7r^2)/(r+3) ?

The integral of (7r^2)/(r+3) is 7(((r+3)^2)/2-6(r+3)+9ln|r+3|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7r^2)/(r+3)

\int \frac{7 r ^{2}}{r + 3} d r

7 (\frac{( r + 3 ) ^{2}}{2} - 6 (r + 3) + 9 ln ∣ r + 3 ∣) + C

Solution steps

\int \frac{7 r ^{2}}{r + 3} d r

Take the constant out:

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

= 7 \cdot \int \frac{r ^{2}}{r + 3} d r

Apply u-substitution

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

= 7 \cdot \int \frac{( u - 3 ) ^{2}}{u} d u

Expand

\frac{( u - 3 ) ^{2}}{u} : u - 6 + \frac{9}{u}

= 7 \cdot \int u - 6 + \frac{9}{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

= 7 (\int u d u - \int 6 d u + \int \frac{9}{u} d u)

\int u d u = \frac{u ^{2}}{2}

\int 6 d u = 6 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= 7 (\frac{u ^{2}}{2} - 6 u + 9 ln ∣ u ∣)

Substitute back

u = r + 3

= 7 (\frac{( r + 3 ) ^{2}}{2} - 6 (r + 3) + 9 ln ∣ r + 3 ∣)

Add a constant to the solution

= 7 (\frac{( r + 3 ) ^{2}}{2} - 6 (r + 3) + 9 ln ∣ r + 3 ∣) + C

2 x^{2} - 9 x + 10, at x = 1

\int \frac{( 1 - 2 y )}{y} d y

\int e^{2 x} \cdot sin (2 x) d x

f (x) = x^{2}, at x = - 5

(\frac{1}{s}) (\frac{10}{s + 5})

What is the integral of (7r^2)/(r+3) ?
The integral of (7r^2)/(r+3) is 7(((r+3)^2)/2-6(r+3)+9ln|r+3|)+C