int sqrt(1-x^2)/x^4 dx Find the indefinite integral

Given ,
int sqrt(1-x^2)/x^4 dx
By applying Integration by parts we can solve the given integral
so,
let u= sqrt(1-x^2) ,v' = (1/x^4)
=> u' = (sqrt(1-x^2) )'
=> =d/dx (sqrt(1-x^2))
let t=1-x^2
so,
d/dx (sqrt(1-x^2))
=d/dx (sqrt(t))
= d/(dt) sqrt(t) * d/dx (t)        [as d/dx f(t) = d/(dt) f(t) (dt)/dx ]
=  [(1/2)t^((1/2)-1) ]*(d/dx (1-x^2))
=  [(1/2)t^(-1/2)]*(-2x)
=[1/(2sqrt(1-x^2 ))]*(-2x)
=-x/sqrt(1-x^2)
so,  u' = -x/sqrt(1-x^2) and as  v'=(1/x^4) so
v = int 1/x^4 dx
= int x^(-4) dx
= (x^(-4+1))/(-4+1)
=(x^(-3))/(-3)
= -(1/(3x^3))
 
so , let us see the values altogether.
u= sqrt(1-x^2) ,u' = -x/sqrt(1-x^2) and v' = (1/x^4) ,v=-(1/(3x^3))
 
Now ,by applying the integration by parts int uv' is given as
 int uv' = uv - int u'v
then,
int sqrt(1-x^2)/x^4 dx
= (sqrt(1-x^2)) (-(1/(3x^3))) - int (-x/sqrt(1-x^2))(-(1/(3x^3))) dx
= (sqrt(1-x^2)) (-(1/(3x^3))) -(- int (-x/sqrt(1-x^2))((1/(3x^3))) dx)
= (sqrt(1-x^2)) (-(1/(3x^3))) - int (x/sqrt(1-x^2))((1/(3x^3))) dx
= -(sqrt(1-x^2)) ((1/(3x^3))) - int (x/sqrt(1-x^2))((1/(3x^3))) dx-----(1)
 
Now let us solve ,
int (x/sqrt(1-x^2))((1/(3x^3))) dx
=int (1/sqrt(1-x^2))((1/(3x^2))) dx
=(1/3)int (1/sqrt(1-x^2))((1/(x^2))) dx
=(1/3)int (1/((x^2)sqrt(1-x^2))) dx
This integral can be solve by using the Trigonometric substitution(Trig substitution)
 
when the integrals containing sqrt(a-bx^2)then we have to take x=sqrt(a/b) sin(t)to solve the integral easily
 
so here , For
(1/3)int (1/((x^2)sqrt(1-x^2))) dx------(2)
x is given as
x= sqrt(1/1) sin(t) = sin(t)
as x= sin(t)
=>  dx= cos(t) dt
now substituting the value of x in (2) we get
(1/3)int (1/((x^2)sqrt(1-x^2))) dx
=(1/3)int (1/(((sin(t))^2)sqrt(1-(sin(t))^2))) (cos(t) dt)
= (1/3)int (1/(((sin(t))^2)sqrt(cos(t))^2))) (cos(t) dt)
=(1/3)int (1/(((sin(t))^2)*(cos(t)))) (cos(t) dt)
=(1/3)int 1/(((sin(t))^2)) dt
=(1/3)int (csc(t))^2 dt     
= (-1/3) cot(t) +c
= (-1/3) cot(arcsin(x)) +c ---(3)   
[since x= sin(t) =>      t= arcsin(x)]
 
Now substituting (3) in (1) we get
(1) =>
-(sqrt(1-x^2)) ((1/(3x^3))) - int (x/sqrt(1-x^2))((1/(3x^3))) dx
=-(sqrt(1-x^2)) ((1/(3x^3))) - ((-1/3) cot(arcsin(x)) +c)
=-(sqrt(1-x^2)) ((1/(3x^3)))+(1/3) cot(arcsin(x)) +c
=(1/3) cot(arcsin(x))- (((sqrt(1-x^2))/(3x^3))) +c----(4)
cot(t) in terms of sin(t) can be given as follows
cot(t) = cos(t)/sin(t) = sqrt(1-(sin(t))^2)/sin(t)
so,
cot(arcsin(x)) = sqrt(1-(sin(arcsin(x)))^2)/sin(arcsin(x)) = sqrt(1-x^2)/x
substituting the above in (4) we get
(1/3) cot(arcsin(x))- (((sqrt(1-x^2))/(3x^3))) +c
=(1/3) (sqrt(1-x^2)/x)- (((sqrt(1-x^2))/(3x^3))) +c
=(sqrt(1-x^2)/(3x))- (((sqrt(1-x^2))/(3x^3))) +c
so,
int sqrt(1-x^2)/x^4 dx
=(sqrt(1-x^2)/(3x))- sqrt(1-x^2)/(3x^3)+c