What is the integral of (8x)/((1-x^2)^4) ?

The integral of (8x)/((1-x^2)^4) is 4/(3(-x^2+1)^3)+C

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integral of (8x)/((1-x^2)^4)

\int \frac{8 x}{( 1 - x ^{2} ) ^{4}} d x

\frac{4}{3 ( - x ^{2} + 1 ) ^{3}} + C

Solution steps

\int \frac{8 x}{( 1 - x ^{2} ) ^{4}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{( 1 - x ^{2} ) ^{4}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{4}} d u

= 8 \cdot \int - \frac{1}{2 u ^{4}} d u

Take the constant out:

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

= 8 (- \frac{1}{2} \cdot \int \frac{1}{u ^{4}} d u)

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 8 (- \frac{1}{2} (- \frac{1}{3 u ^{3}}))

Substitute back

u = 1 - x^{2}

= 8 (- \frac{1}{2} (- \frac{1}{3 ( 1 - x ^{2} ) ^{3}}))

Simplify

8 (- \frac{1}{2} (- \frac{1}{3 ( 1 - x ^{2} ) ^{3}})) : \frac{4}{3 ( - x ^{2} + 1 ) ^{3}}

= \frac{4}{3 ( - x ^{2} + 1 ) ^{3}}

Add a constant to the solution

= \frac{4}{3 ( - x ^{2} + 1 ) ^{3}} + C

\frac{\partial}{\partial z} (\frac{4 x - 4 z}{y + 4 z})

y y^{^{'}} = x - 2 x^{3}

\frac{d}{d x} (a^{7} + (cos (x))^{7})

x = y^{2} + 4 y - 22, x = 3 y + 8

x - tan (x)

What is the integral of (8x)/((1-x^2)^4) ?
The integral of (8x)/((1-x^2)^4) is 4/(3(-x^2+1)^3)+C