Question

How do i integrate x^4/(x^2+1)^3 ?

Answer 1

`I=1/8[3tan^-1x-(5x^3+3x)/((1+x^2)^2)]`

Explanation:

Here,

`I=intx^4/(x^2+1)^3dx`

Let, `x=tant=>dx=sec^2tdt`

`andx^2+1=tan^2t+1=sec^2t`

So,`I=inttan^4t/(sec^2t)^3sec^2tdt`

#=inttan^4t/sec^4tdt=int(sin^4t/cos^4t)/(1/cos^4t)dt=intsin^4tdt #

`=int(sin^2t)^2dt=int((1-cos2t)/2)^2dt`

`=1/4int(1-2cos2t+cos^2 2t)dt`

`=1/4int(1-2cos2t+(1+cos4t)/2)dt`

`=1/8int(3-4cos2t+cos4t)dt`

`=1/8[3t-(4sin2t)/2+(sin4t)/4]+c`

`=1/8[3t-2((2tant)/(1+tan^2t))+1/4(2sin2tcos2t)]+c`

#=1/8[3t-(4tant)/(1+tan^2t)+1/2((2tant)/(1+tan^2t))((1- tan^2t)/(1+tan^2t))]#

Substituting back `x=tant=>t=tan^-1x`,we get

`=1/8[3tan^-1x-(4x)/(1+x^2)+x/(1+x^2)xx(1-x^2)/(1+x^2)]+c`

`I=1/8[3tan^-1x-((4x(1+x^2)-x(1-x^2))/(1+x^2)^2)]`

`I=1/8[3tan^-1x-(5x^3+3x)/((1+x^2)^2)]`

Answer 2

`int (x^4dx)/(1+x^2)^3 = (3arctanx)/8 -(5x^3+3x)/(8(1+x^2)^2) +C`

Explanation:

Let:

`x= tant`

with `t in (-pi/2,pi/2)`

`dx = sec^2t dt`

`1+x^2 = 1+tan^2t = sec^2t`

Then:

`int (x^4dx)/(1+x^2)^3 = int ( tan^4t sec^2t dt)/sec^6t`

`int (x^4dx)/(1+x^2)^3 = int tan^4t /sec^4t dt`

and as:

`tant/sect = (sint /cost )/(1/cost) = sint`

then:

`int (x^4dx)/(1+x^2)^3 = int sin^4t dt`

Write the integrand as:

`sin^4t = sin^2t (1-cos^2t) = sin^2t- sin^2tcos^2t`

`sin^4t = 1/2- (cos 2t)/2 - (sin^2 2t)/4`

`sin^4t = 1/2- (cos 2t)/2 - (1-cos4t)/8`

`sin^4t = 3/8 - (cos 2t)/2 +(cos4t)/8`

Then using the linearity of the integral:

`int (x^4dx)/(1+x^2)^3 = 3/8 int dt - 1/2 int cos2tdt +1/8 int cos4t dt`

`int (x^4dx)/(1+x^2)^3 = (3t)/8 - (sin2t)/4 + (sin4t)/32 +C`

To undo the substitution use the trigonometric formulas:

`sin 4t = 2 sin 2t cos 2t`

to have:

`int (x^4dx)/(1+x^2)^3 = (3t)/8 - (sin2t)/4 + (sin2tcos2t)/16 +C`

and then the parametric formulas:

`sin 2t = (2tant)/(1+tan^2t) = (2x)/(1+x^2)`

`cos 2t = (1-tan^2t)/(1+tan^2t) = (1-x^2)/(1+x^2)`

so:

`int (x^4dx)/(1+x^2)^3 = (3arctanx)/8 -x/(2(1+x^2)) + (x(1-x^2))/(8(1+x^2)^2) +C`

and simplifying:

`int (x^4dx)/(1+x^2)^3 = (3arctanx)/8 -(4x(1+x^2) - x(1-x^2))/(8(1+x^2)^2) +C`

`int (x^4dx)/(1+x^2)^3 = (3arctanx)/8 -(5x^3+3x)/(8(1+x^2)^2) +C`