Question

`int(4x^2-x+12)/(x^3+4x)dx`?

Answer 1

`3ln|x|+1/2ln(x^2+4)-1/2arc tan(x/2)+C, or,`

`ln|x^3*sqrt(x^2+4)|-1/2arc tan(x/2)+C`.

Explanation:

Let, `I=int(4x^2-x+12)/(x^3+4x)dx=int(4x^2-x+12)/{x(x^2+4)}dx`.

We will use Heaviside's Method to decompose the

integrand `(4x^2-x+12)/{x(x^2+4)}` into Partial Fraction :

`color(red)(4x^2-x+12)/{color(blue)(x)(color(red)(x^2+4))}=A/x+(Bx+C)/(x^2+4)," say for some "A,B,C in RR`.

Then, we have, `A=[color(red)((4x^2-x+12)/(x^2+4))]_(color(blue)(x=0))=12/4=3`.

Sub.ing `A=3` above, we have,

`(4x^2-x+12)/{x(x^2+4)}-3/x=(Bx+C)/{x(x^2+4)}, i.e.,`

`(Bx+C)/{x(x^2+4)}={(4x^2-x+12)-3(x^2+4)}/{x(x^2+4)}`,

`=(x^2-x)/{x(x^2+4)}`.

`rArr (Bx+C)/(x^2+4)=(x-1)/(x^2+4)," giving, "B=1, C=-1`.

Altogether, `(4x^2-x+12)/{x(x^2+4)}=3/x+(x-1)/(x^2+4)`.

`:. I=int3/xdx+int(x-1)/(x^2+4)dx`,

`=3ln|x|+intx/(x^2+4)dx-int1/(x^2+2^2)dx`,

`=3ln|x|+1/2int(2x)/(x^2+4)dx-1/2arc tan(x/2)`,

`=3ln|x|+1/2int{d/dx(x^2+4)}/(x^2+4)dx-1/2arc tan(x/2)`.

`rArr I=3ln|x|+1/2ln(x^2+4)-1/2arc tan(x/2)+C, or,`

`I=ln|x^3*sqrt(x^2+4)|-1/2arc tan(x/2)+C`.

Feel & Spread the Joy of Maths.!

Answer 2

The answer is `=3ln(|x|)+1/2ln(x^2+4)-1/2arctan (x/2)+C`

Explanation:

The degree of the denominator is `>` than the degree of the numerator

The denominator is

`=(x^3+4x)=(x(x^2+4))`

Perform the decomposition into partial fractions

`(4x^2-x+12)/(x^3+4x)=(4x^2-x+12)/(x(x^2+4))`

`=A/(x)+(Bx+C)/(x^2+4)`

`=(A(x^2+4)+x(Bx+C))/(x(x^2+4))`

The denominators are the same, compare the numerators

`4x^2-x+12=A(x^2+4)+x(Bx+C)`

Let `x=0`, `=>`, `12=4A`, `=>`, `A=3`

Coefficients of `x^2`

`4=A+B`, `=>`, `B=4-A=4-3=1`

Coefficients of `x`

`-1=C`

Therefore,

`(4x^2-x+12)/(x^3+4x)=3/(x)+(x-1)/(x^2+4)`

So,
`int((4x^2-x+12dx)/(x^3+4x)=int(3dx)/(x)+int((x-1)dx)/(x^2+4)`

`=3ln(|x|)+int(xdx)/(x^2+4)-int(dx)/(x^2+4)`

`=3ln(|x|)+1/2ln(x^2+4)-1/4int(dx)/((x/2)^2+1)`

`=3ln(|x|)+1/2ln(x^2+4)-1/2arctan (x/2)+C`