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

Question

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!

Answer 1

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 )

Answer 2

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

Answer 3

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

Answer 4

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