What is the integral of (-2x^2)/(9-x^2) ?

The integral of (-2x^2)/(9-x^2) is -6(1/2 ln| x/3+1|-1/2 ln| x/3-1|-x/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-2x^2)/(9-x^2)

\int \frac{- 2 x ^{2}}{9 - x ^{2}} d x

- 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) + C

Solution steps

\int \frac{- 2 x ^{2}}{9 - x ^{2}} d x

Take the constant out:

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

= - 2 \cdot \int \frac{x ^{2}}{9 - x ^{2}} d x

Apply Integral Substitution

= - 2 \cdot \int \frac{3 u ^{2}}{1 - u ^{2}} d u

Take the constant out:

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

= - 2 \cdot 3 \cdot \int \frac{u ^{2}}{1 - u ^{2}} d u

\frac{u ^{2}}{1 - u ^{2}} = \frac{1}{1 - u ^{2}} - 1

= - 2 \cdot 3 \cdot \int \frac{1}{1 - u ^{2}} - 1 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

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

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= - 2 \cdot 3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = \frac{x}{3}

= - 2 \cdot 3 (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2} - \frac{x}{3})

Simplify

- 2 \cdot 3 (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2} - \frac{x}{3}) : - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})

= - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})

Add a constant to the solution

= - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) + C

\int (3 x^{4} - 2 x^{3} + 5) d x

\int \frac{2 x}{16 x ^{2} - 25} d x

\int u (u^{2} + 3 u + 7) d u

\int x^{8} (x^{9} - 10)^{4} d x

What is the integral of (-2x^2)/(9-x^2) ?
The integral of (-2x^2)/(9-x^2) is -6(1/2 ln| x/3+1|-1/2 ln| x/3-1|-x/3)+C