What is the integral of (3r^2)/(r+9) ?

The integral of (3r^2)/(r+9) is 3(((r+9)^2)/2-18(r+9)+81ln|r+9|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3r^2)/(r+9)

\int \frac{3 r ^{2}}{r + 9} d r

3 (\frac{( r + 9 ) ^{2}}{2} - 18 (r + 9) + 81 ln ∣ r + 9 ∣) + C

Solution steps

\int \frac{3 r ^{2}}{r + 9} d r

Take the constant out:

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

= 3 \cdot \int \frac{r ^{2}}{r + 9} d r

Apply u-substitution

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

= 3 \cdot \int \frac{( u - 9 ) ^{2}}{u} d u

Expand

\frac{( u - 9 ) ^{2}}{u} : u - 18 + \frac{81}{u}

= 3 \cdot \int u - 18 + \frac{81}{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

= 3 (\int u d u - \int 18 d u + \int \frac{81}{u} d u)

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

\int 18 d u = 18 u

\int \frac{81}{u} d u = 81 ln ∣ u ∣

= 3 (\frac{u ^{2}}{2} - 18 u + 81 ln ∣ u ∣)

Substitute back

u = r + 9

= 3 (\frac{( r + 9 ) ^{2}}{2} - 18 (r + 9) + 81 ln ∣ r + 9 ∣)

Add a constant to the solution

= 3 (\frac{( r + 9 ) ^{2}}{2} - 18 (r + 9) + 81 ln ∣ r + 9 ∣) + C

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

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

\int x^{2} cot (x^{3}) d x

\frac{\partial}{\partial x} (\frac{3 x}{4 y})

3 x + \frac{5}{x}

What is the integral of (3r^2)/(r+9) ?
The integral of (3r^2)/(r+9) is 3(((r+9)^2)/2-18(r+9)+81ln|r+9|)+C