integral of y^3/(y^4 +1) dy

I don't get how you get (1/4) ln(y^4 +1) out of this. I thought you would do the quotient rule. but is that the same for integrals as it is the derivatice?

help!

Substitution rule:

y ^ 4 + 1 = t

4 y ^ 3 dy = dt Divide both sides by 4

y ^ 3 dy = dt / 4

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

integral of ( dt / 4 ) / t =

( 1 / 4 ) integral of dt / t =

( 1 / 4 ) ln ( t ) =

( 1 / 4 ) ln ( y ^ 4 + 1 )

thanks so much! can you help me understand this one as well?

I have the integral from 0 to 2-2x of integrate (2-2x-y)dy... I thought you just take the integral with respect to y so it would be -y^2/2? (plugging the bounds it that would be -(2-2x)^2/2 but it comes out to be (2-2x)^2 - (2-2x)^2/2

integral of ( 2 - 2 x - y ) dy =

2 y - 2 x y - y ^ 2 / 2

for y = 0

2 y - 2 x y - y ^ 2 / 2 =

2 * 0 - 2 x * 0 - 0 ^ 2 / 2 = 0

for y = 2 - 2 x

2 ( 2 - 2 x ) - 2 x ( 2 - 2 x ) - ( 2 - 2 x ) ^ 2 / 2 =

( 2 - 2 x ) ( 2 - 2 x ) - ( 2 - 2 x ) ^ 2 / 2 =

( 2 - 2 x ) ^ 2 - ( 2 - 2 x ) ^ 2 / 2 =

( 2 - 2 x ) ^ 2 / 2 =

[ 2 ( 1 - x ) ] ^ 2 / 2 =

2 ^ 2 ( 1 - x ) ^ 2 / 2 =

4 ( 1 - x ) ^ 2 / 2 =

2 ( 1 - x ) ^ 2

integral from 0 to 2 - 2 x of ( 2 - 2 x - y ) dy =

2 ( 1 - x ) ^ 2 - 0 = 2 ( 1 - x ) ^ 2

By the way :

( 1 - x ) ^ 2 =

[ ( - 1 ) ( x - 1 ) ] ^ 2 =

( - 1 ) ^ 2 ( x - 1 ) ^ 2 =

1 * ( x - 1 ) ^ 2 =

( x - 1 ) ^ 2

integral from 0 to 2 - 2 x of ( 2 - 2 x - y ) dy =

2 ( 1 - x ) ^ 2 = 2 ( x - 1 ) ^ 2

To find the integral of the function y^3/(y^4 + 1) dy, we can use a u-substitution. Let's go through the steps:

1. Start with the original integral: ∫ y^3/(y^4 + 1) dy.

2. Make a substitution: Let u = y^4 + 1. This means du = 4y^3 dy.

3. Rearrange the equation to solve for dy: dy = du/(4y^3).

4. Substitute the values of u and dy into the original integral: ∫ (y^3/(y^4 + 1)) dy = ∫ ((1/4)*(du/y^3)).

5. Simplify the integral: ∫ (1/4) * (du/y^3).

6. Rewrite the integral: (1/4) ∫ (1/y^3) du.

7. Integrate the new expression: (1/4) ∫ (u^(-3)) du.

8. Apply the power rule of integration: (1/4) * (u^(-3 + 1))/(-3 + 1) + C.

9. Simplify the expression: (1/4) * (u^(-2))/(-2) + C.

10. Substitute the original value of u back: (1/4) * (y^4 + 1)^(-2)/(-2) + C.

11. Simplify further: -1/8 * (y^4 + 1)^(-2) + C.

So, the final answer to the integral is -1/8 * (y^4 + 1)^(-2) + C.

The result of (1/4) ln(y^4 + 1) you mentioned is incorrect. It seems like an error may have occurred during the integration process. The correct answer is -1/8 * (y^4 + 1)^(-2) + C.