Question

integrate 3x^2+1/(x^2-1)^3​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1 )^{3} } dx.$

As we know that,

We can write this equation as,

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1^{2}{} )^{3} }dx$

As we know that,

Formula of :

⇒ (x² - y²) = (x + y)(x - y).

Using this formula in equation, we get.

$\sf \implies \displaystyle \int\dfrac{3x^{2} + 1}{[(x + 1)(x - 1)]^{3} } dx$

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x + 1)^{3}(x - 1)^{3} } dx.$

As we know that,

Expand the equation of (x + 1)³ and (x - 1)³, we get.

⇒ (x + 1)³ = x³ + 3x² + 3x + 1.

⇒ (x - 1)³ = x³ - 3x² + 3x - 1.

Now we add both equation, we get.

⇒ (x + 1)³ + (x - 1)³ = x³ + 3x² + 3x + 1 + x³ - 3x² + 3x - 1.

⇒ (x + 1)³ + (x - 1)³ = 2x³ + 6x.

This equation is not satisfying.

Now we subtract the equation, we get.

⇒ (x + 1)³ - (x - 1)³ = x³ + 3x² + 3x + 1 - [x³ - 3x² + 3x - 1].

⇒ (x + 1)³ - (x - 1)³ = x³ + 3x² + 3x + 1 - x³ + 3x² - 3x + 1.

⇒ (x + 1)³ - (x - 1)³ = 6x² + 2.

⇒ (x + 1)³ - (x - 1)³ = 2(3x² + 1).

As we can see that,

Multiply and divide numerator and denominator by 2, we get.

$\sf \implies \displaystyle \dfrac{1}{2} \int \dfrac{(x + 1)^{3} - (x - 1)^{3} }{(x + 1)^{3} (x - 1)^{3} } dx$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg(\dfrac{(x + 1)^{3} }{(x + 1)^{3} (x - 1)^{3}} \ - \ \dfrac{(x - 1)^{3} }{(x + 1)^{3} (x - 1)^{3} } \bigg)dx.$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg(\dfrac{1}{(x - 1)^{3} } \ - \ \dfrac{1}{(x + 1)^{3} } \bigg)dx.$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg[(x - 1)^{-3} \ - \ (x + 1)^{-3} \bigg]dx.$

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

Using this formula in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{2} \bigg[\dfrac{(x - 1)^{-3 + 1} }{- 3 + 1} \ - \ \dfrac{(x + 1)^{-3 + 1} }{-3 + 1} \bigg] + c.$

$\sf \implies \displaystyle \dfrac{1}{2} \bigg[ \dfrac{(x - 1)^{-2} }{-2} \ - \ \dfrac{(x + 1)^{-2} }{-2} \bigg] + c.$

$\sf \implies \displaystyle \dfrac{1}{4} \bigg[ \dfrac{1}{(x + 1)^{2} } \ - \ \dfrac{1}{(x - 1)^{2} } \bigg] + c.$

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1)^{3} } dx = \dfrac{1}{4} \bigg[ \dfrac{1}{(x + 1)^{2} } \ - \ \dfrac{1}{(x - 1)^{2} } \bigg] + c.$

                                                                                                                       

MORE INFORMATION.

Standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, ( k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, ( n ≠ - 1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.

Answer 2

$\huge\mathcal{\pink{A}}\huge\mathcal{\purple{N}}\huge\mathcal{\green{S}}\huge\mathcal{\blue{W}}\huge\mathcal{\red{E}}\huge\mathcal{R}$

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1 )^{3} } dx.$

As we know that,

We can write this equation as,

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1^{2}{} )^{3} }dx$

As we know that,

Formula of :

⇒ (x² - y²) = (x + y)(x - y).

Using this formula in equation, we get.

$\sf \implies \displaystyle \int\dfrac{3x^{2} + 1}{[(x + 1)(x - 1)]^{3} } dx$

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x + 1)^{3}(x - 1)^{3} } dx.$

As we know that,

Expand the equation of (x + 1)³ and (x - 1)³, we get.

⇒ (x + 1)³ = x³ + 3x² + 3x + 1.

⇒ (x - 1)³ = x³ - 3x² + 3x - 1.

Now we add both equation, we get.

⇒ (x + 1)³ + (x - 1)³ = x³ + 3x² + 3x + 1 + x³ - 3x² + 3x - 1.

⇒ (x + 1)³ + (x - 1)³ = 2x³ + 6x.

This equation is not satisfying.

Now we subtract the equation, we get.

⇒ (x + 1)³ - (x - 1)³ = x³ + 3x² + 3x + 1 - [x³ - 3x² + 3x - 1].

⇒ (x + 1)³ - (x - 1)³ = x³ + 3x² + 3x + 1 - x³ + 3x² - 3x + 1.

⇒ (x + 1)³ - (x - 1)³ = 6x² + 2.

⇒ (x + 1)³ - (x - 1)³ = 2(3x² + 1).

As we can see that,

Multiply and divide numerator and denominator by 2, we get.

$\sf \implies \displaystyle \dfrac{1}{2} \int \dfrac{(x + 1)^{3} - (x - 1)^{3} }{(x + 1)^{3} (x - 1)^{3} } dx$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg(\dfrac{(x + 1)^{3} }{(x + 1)^{3} (x - 1)^{3}} \ - \ \dfrac{(x - 1)^{3} }{(x + 1)^{3} (x - 1)^{3} } \bigg)dx.$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg(\dfrac{1}{(x - 1)^{3} } \ - \ \dfrac{1}{(x + 1)^{3} } \bigg)dx.$

$\sf \implies \displaystyle \dfrac{1}{2} \int \bigg[(x - 1)^{-3} \ - \ (x + 1)^{-3} \bigg]dx.$

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

Using this formula in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{2} \bigg[\dfrac{(x - 1)^{-3 + 1} }{- 3 + 1} \ - \ \dfrac{(x + 1)^{-3 + 1} }{-3 + 1} \bigg] + c.$

$\sf \implies \displaystyle \dfrac{1}{2} \bigg[ \dfrac{(x - 1)^{-2} }{-2} \ - \ \dfrac{(x + 1)^{-2} }{-2} \bigg] + c.$

$\sf \implies \displaystyle \dfrac{1}{4} \bigg[ \dfrac{1}{(x + 1)^{2} } \ - \ \dfrac{1}{(x - 1)^{2} } \bigg] + c.$

$\sf \implies \displaystyle \int \dfrac{3x^{2} + 1}{(x^{2} - 1)^{3} } dx = \dfrac{1}{4} \bigg[ \dfrac{1}{(x + 1)^{2} } \ - \ \dfrac{1}{(x - 1)^{2} } \bigg] + c.$

                                                                                                                       

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