Solve the integral of y=x^3/(x^4+1)^5 .
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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]
=> (-1/16)/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
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
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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