Last one for me.

Question

Evaluate (3x^2)/sqrt(1-x^6) dx

I changed sqrt (1-x^6) into 1^(1/2)- x^(3)
and let u=1^(1/2)- x^(3)

then I got du/dx=3x^2 and subbed dx=du/3x^2 into the equation and got x-(x^4/4) but its not right.

Options

7x^3/(3sqrt(1-x^7) + C
cos-1(3x) + C
sin-1(x^3) + C
The antiderivative cannot be found

Answer 1

sqrt (1-x^6) into 1^(1/2)- x^(3)
I do not think so

I will try the easy way
u = x^3
du = 3 x^2 dx

so I have
du/ sqrt (1-u^2)

Answer 2

Wait, but how can you let u=x^3 when there is not an x^3 in the equation

Answer 3

sin^-1 u + c
so
sin^-1 x^3 + c

Answer 4

Why should I not let u = anything I choose ? LOL
I see an x^6 and an x^2 so I tried x^3

Answer 5

Okay I didn't know you could do that haha

Answer 6

By the way, you tried x^3 as well
:..... sqrt (1-x^6) into 1^(1/2)- x^(3) ..."

Answer 7

It is just that
sqrt (a^2-b^2)
is not
a - b