Find the indefinate integral. (1-3 x^(1/2) ) / x^2?

Question

it's written as 1 - 3 square root of x over x^2

Answer 1

=1/x^2 - 3/[x^(3/2)]
int = -1/x + 3/√x (1/0.5) +c
= 6/√x -1/x +c

Answer 2

a + b + c = 0 a² + b² + c² = a million a³ + b³ + c³ = 0 (a + b + c)³ + 2(a³ + b³ + c³) = 0³ + 2(0) (a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 6abc) + 2(a³ + b³ + c³) = 0 3 (a²b + a²c + b²a + b²c + c²a + c²b) + 3 (a³ + b³ + c³) + 6abc = 0 3 (a³ + a²b + a²c + b²a + b³ + b²c + c²a + c²b + c³) + 6abc = 0 3 (a² (a + b + c) + b² (a + b + c) + c² (a + b + c)) + 6abc = 0 3 (a + b + c) (a² + b² + c²) + 6abc = 0 3 (0) (a million) + 6abc = 0 6 abc = 0 one in all a, b, c could be = 0 with out loss of generality, permit c = 0 Then we get: a + b = 0 a² + b² = a million a³ + b³ = 0 From first equation b = ?a third equation turns into: a³ + (?a)³ = 0 a³ ? a³ = 0 that's actual for all a 2nd equation turns into: a² + (?a)² = a million 2a² = a million a² = a million/2 a = ± a million/?2 whilst a = a million/?2, b = ?a million/?2 whilst a = ?a million/?2, b = a million/?2 So the three numbers, in some order, are: 0 a million/?2 ?a million/?2 Six accessible strategies: x? = 0, x? = a million/?2, x? = ?a million/?2 x? = 0, x? = ?a million/?2, x? = a million/?2 x? = a million/?2, x? = 0, x? = ?a million/?2 x? = a million/?2, x? = ?a million/?2, x? = 0 x? = ?a million/?2, x? = 0, x? = a million/?2 x? = ?a million/?2, x? = a million/?2, x? = 0 M?thm?m

Answer 3

it would probobly be easier if you split it up to:

1/x^2-3x^(1/2)/x^2

which is the same as 1/x^2-3/x(3/2)

that makes taking the integral a lot easier...(use basic formula x^n+1/n+1)

=-1/x+6/x(1/2) + C

Answer 4

I = ∫ x^(-2) - 3.x^(- 3/2) dx
I = - x^(-1) + 6.x^(-1/2) + C
I = - 1/x + 6 /√x + C