Answer to Question #185650 in Calculus for phyroe

Question

Integration by Parts Fractions

1.) ∫4dx/(x^4-1)

2.) ∫8dx/x(x^2+2)^2

Answer 1

1) ∫ 4 dx / (x^4 - 1)

We know that x⁴ - 1 = (x² -1) * (x² +1)

Also (x² +1) - (x² -1) = 2

So the above integral can be written as

2 ∫ [(x²+1) - (x²-1)] dx / (x²+1) * (x²-1)

2 ∫ [1/(x²-1) - 1/(x²+1) ] dx

2 ln(x-1/x+1) - 2 arctan (x) + c [Since ∫ dx/x² - a² = (1/2a ) ln(x-a/ x+a) + c

and ∫ dx/a²+x² = (1/a ) arctan(x/a) + c ]

2) ∫ 8 dx/ x(x²+2)²

The above integral can be expressed as a sum of partial fractions as shown below

8 / x(x²+2)² = A/x + Bx/(x²+2) + Cx/ (x²+2)²

On comparing the coefficients in LHS and RHS and solving them we have

A = 2

B = -2

C = -4

Now the above integral takes the form

∫ [ 2/x - 2x/(x²+2) - 4x/ (x²+2)² ] dx

Let I₁ = ∫ 2 dx/x

I₂ = ∫ 2x dx/(x²+2)

I₃ = ∫ 4x dx/(x²+2)² ................(1)

I₁ = 2 ln(x)

For I₂ ,

Let x²+2 = t

2x dx = dt

∴ I₂ = ∫ dt / t

= ln(t)

= ln( x²+2 ) ....................(2)

For I₃ ,

Let x²+2 = u

2x dx = du

∴ I₃ = 2 ∫ du / u²

= - 2 /u

= -2/ (x²+2) ................(3)

∴ ∫ 8 dx/ x(x²+2)² = I₁ - I₂ - I₃

= 2 ln(x) - ln (x² + 2) + 2 / (x² + 2) + c

= ln[ x² / (x² + 2)] + 2 / (x²+2) + c