Find antiderivative of (4x)/(1-x^4)^0.5

[hndalama]

integral of (4x) / (1-x⁴)^0.5 dx

I have a worked solution to the question but I'd like to know what is wrong with the method I used.
I used u substitution
u = 1- x⁴
du = -4x³ dx
-1/(4)^1/3 du = x dx

integral of: -4 / [(4)^1/3(u)^0.5]

antiderivative = (-8u^0.5) /(4)^1/3 = [-8(1-x⁴)^0.5] / (4)^​1/3

 

[Deleted member 4993]

integral of (4x) / (1-x⁴)^0.5 dx

I have a worked solution to the question but I'd like to know what is wrong with the method I used.
I used u substitution
u = 1- x⁴
du = -4x³ dx
-1/(4)^1/3 du = x dx ← How did you get that?


integral of: -4 / [(4)^1/3(u)^0.5]

antiderivative = (-8u^0.5) /(4)^1/3 = [-8(1-x⁴)^0.5] / (4)^​1/3

You should plan to use trigonometric substitution.

x^2 = sin(u)

and continue......

 

[ksdhart2]

Yet another way is to let u = x^2, leaving a denominator of sqrt(1-u^2), which is a form that looks like one of the known integrals of inverse trig functions.

 

[hndalama]

You should plan to use trigonometric substitution.

x^2 = sin(u)

and continue......

I got -1/(4)^1/3 du = x dx by first divide both sides by -4 to get (-1/4)du =x³ dx then I took the cube root of both sides to get (-1/4)^1/3du= x dx.
is something wrong with that?

 

[Deleted member 4993]

I got -1/(4)^1/3 du = x dx by first divide both sides by -4 to get (-1/4)du =x³ dx then I took the cube root of both sides to get (-1/4)^1/3du= x dx.
is something wrong with that? .........Yes

You cannot do that with differential elements!

 

[HallsofIvy]

integral of (4x) / (1-x⁴)^0.5 dx

I have a worked solution to the question but I'd like to know what is wrong with the method I used.
I used u substitution
u = 1- x⁴
du = -4x³ dx
-1/(4)^1/3 du = x dx

What??? How in the world did you get that last part? From du= -4x³ dx you get -(1/4)du= x³dx. You can't just take the third root of "-4x³" without taking the third root of the rest of the equation! And third roots of "du" and "dx" make no sense!

integral of: -4 / [(4)^1/3(u)^0.5]

antiderivative = (-8u^0.5) /(4)^1/3 = [-8(1-x⁴)^0.5] / (4)^​1/3