Integrals: dy/dx = (4x^3 + 1)/x^2, dy/dx = (x + 4)/sqrt[x],

[1141]

Can someone please explain to me how to find the integrals of the following equations:

a.) dy/dx = (4x[sup:1wz7due6]3[/sup:1wz7due6] + 1) / x[sup:1wz7due6]2[/sup:1wz7due6]

b.) dy/dx = (x + 4) / ?x

c.) dy/dx = (?x +5)[sup:1wz7due6]2[/sup:1wz7due6]

d.) dy/dx = (?x + 5) / ?x

 

[arthur ohlsten]

Re: Integrals

a)
dy/dx=[4x^3 + 1] / x^2
improper fraction, divide
dy = [4x + x^-2] dx
integrate
y= 4x^2/2 +x^-1/[-1] +c
y=2x^2 - 1/x + C answer

b)
dy/dx = [x+4] / x^1/2
divide
dy= [x^1/2 +4x^-1/2] dx
integrate
y= x^3/2 / [3/2] + 4x^1/2 /[1/2] +C
y= [2/3]x^3/2 + 8 x^1/2 + C
y= 2x^1/2[ x/3 +8] +C

c)
dy/dx = [x^1/2 +5 ]^2
square the term
dy = [x +10 x^1/2 +25] dx

y= x^2/2 + 10 x^3/2 / [3/2] +25x +C
y= [x^2]/2 + [20x^3/2]/3 +26x +C

d)
dy/dx = [x^1/2 + 5] /x^1/2
improper fraction
dy = [1 + 5 x^-1/2] dx
y= x + 5x^1/2 /[1/2] +C
y=x + 10 x^1/2 +C

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please check the math and I suggest you take the derivative of the answers to see if the derivative agrees with the original derivative
Arthur

 

[1141]

I checked them all and they all work out. Thanks so much for the help!