What is the integral of (9x)/((1-x^2)^3) ?

The integral of (9x)/((1-x^2)^3) is 9/(4(1-x^2)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9x)/((1-x^2)^3)

\int \frac{9 x}{( 1 - x ^{2} ) ^{3}} d x

\frac{9}{4 ( 1 - x ^{2} ) ^{2}} + C

Solution steps

\int \frac{9 x}{( 1 - x ^{2} ) ^{3}} d x

Take the constant out:

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

= 9 \cdot \int \frac{x}{( 1 - x ^{2} ) ^{3}} d x

Apply u-substitution

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

= 9 \cdot \int - \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 9 (- \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

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

= 9 (- \frac{1}{2} (- \frac{1}{2 u ^{2}}))

Substitute back

u = 1 - x^{2}

= 9 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}}))

Simplify

9 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}})) : \frac{9}{4 ( 1 - x ^{2} ) ^{2}}

= \frac{9}{4 ( 1 - x ^{2} ) ^{2}}

Add a constant to the solution

= \frac{9}{4 ( 1 - x ^{2} ) ^{2}} + C

f (x) = 4 x^{3} 5 x - 1, at x = 1

\int \frac{3 x ^{2}}{( 1 + x ^{3} ) ^{3}} d x

\frac{\partial}{\partial x} (e^{x} y^{2})

\frac{d x}{d t} = \frac{a}{x t}

\int - \frac{1}{2 x + 2} d x

What is the integral of (9x)/((1-x^2)^3) ?
The integral of (9x)/((1-x^2)^3) is 9/(4(1-x^2)^2)+C