Question

integration..........

Answer 1

To add small things is called integrtion
It is use in physics to solve a promblem

Answer 2

Answer:

Step-by-step explanation:

∫dx/(√2x-x^2 )^3

∫dx/(√(x(2-x)))^3

∫dx/x^3/2(√2-x)^3

∫x^-3/2(√2-x)^-3 dx

2x^-1 - 1=t^2

-2x^-2=2t dt

x^-2 dx =-t dt

∫x^-3/2(2-x)^-3/2 dx

∫x^-3/2(x(2x^-1 -1))^-3/2

∫x^-3(2x^- -1) dx

∫x^-2*x^-1(2x^-1 -1 )^-3/2 dx

∫x^-2 dx (2x^-1 -1)^-3/2 x^-1

∫-t dt (t^2)^-3/2 (t^2+1 /2)

-∫t^2+1/ 2t^2

-{∫t^2/2t^2 + ∫ 1/2t^2 }

-{t/2+1/2(t^-1 /-1}

-{t/2 -1/2t }

1/2t-t/2

(1/2√2x^-1 -1 )- (√2x^-1) -1 )/2