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Integrals .Solve the integral of y=x^3/(x^4+1)^5 .

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sodelete | Student, College Freshman | (Level 1) Honors

Posted April 10, 2011 at 11:53 PM via web

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Integrals .

Solve the integral of y=x^3/(x^4+1)^5 .

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 11, 2011 at 12:27 AM (Answer #2)

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We have to find the integral of y = x^3/(x^4+1)^5.

Use substitution to find the integral.

Int [ x^3/(x^4+1)^5 dx]

let x^4 + 1 = y

=> 4*x^3 dx = dy

=> x^3 dx = (1/4) dy

=> (1/4)*Int [ (1/y^5) dy]

=> (1/4)*y^(-4)/(-4)

=> (-1/16)/y^4 + C

substitute y = x^4 + 1

=> (-1/16)/(x^4 + 1)^4 + C

The required integral is (-1/16)/(x^4 + 1)^4 + C

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted April 11, 2011 at 8:01 AM (Answer #3)

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integral of y=x^3/(x^4+1)^5

We'll solve the integral using substitution technique. We'll note x^4 + 1 = t.

We'll differentiate both sides:

4x^3dx = dt

We'll put:

x^3dx = dt/4

We'll write the integral in t:

Int dt/4t^5 = (1/4)Int t^-5dt

(1/4)Int t^-5dt =  (1/4)*t^(-5+1)/(-5+1) +C

Int dt/4t^5 = -(1/16*t^4) + C

We'll substitute t = x^4 + 1

Int x^3*dx/(x^4+1)^5 =  -1/16*(x^4 + 1)^4 + C


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