What is the integral of (7x)/((1-x^2)^3) ?

The integral of (7x)/((1-x^2)^3) is 7/(4(1-x^2)^2)+C

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integral of (7x)/((1-x^2)^3)

\int \frac{7 x}{( 1 - x ^{2} ) ^{3}} d x

\frac{7}{4 ( 1 - x ^{2} ) ^{2}} + C

Solution steps

\int \frac{7 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

= 7 \cdot \int \frac{x}{( 1 - x ^{2} ) ^{3}} d x

Apply u-substitution

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

= 7 \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

= 7 (- \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

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

= 7 (- \frac{1}{2} (- \frac{1}{2 u ^{2}}))

Substitute back

u = 1 - x^{2}

= 7 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}}))

Simplify

7 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}})) : \frac{7}{4 ( 1 - x ^{2} ) ^{2}}

= \frac{7}{4 ( 1 - x ^{2} ) ^{2}}

Add a constant to the solution

= \frac{7}{4 ( 1 - x ^{2} ) ^{2}} + C

x \to π lim (\frac{cos ( x )}{x})

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

y^{^{'}} = 2 x y + 3 y - 4 x - 6

\int 2 x^{3} d x

x \to \infty lim (- e^{x})

What is the integral of (7x)/((1-x^2)^3) ?
The integral of (7x)/((1-x^2)^3) is 7/(4(1-x^2)^2)+C