CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Integrate the rational functions.
3x1(x1)(x2)(x3)dx.
 


Solution

Let 3x1(x1)(x2)(x3)=A(x1)+B(x2)+C(x3)
3x1(x1)(x2)(x3)=A(x2)(x3)+B(x1)(x3)+C(x1)(x2)(x1)(x2)(x3)3x1=A[x25x+6]+B[x24x+3]+C[x23x+2]3x1=x2(A+B+C)+x(5A4B3C)+(6A+3B+2C)

On equating the coefficients of x2, x and constant term on both sides, we get
A+B+C =0 ......(i)
-5A-4B-3C =3 .......(ii)
and 6A+3B+2C=-1 .....(iii)
Form Eq.(i), we get A =-(B+C)
On putting the value of A in Eqs. (ii) and (iii), we get
5((B+C))4B3C=35B+5C4B3C=3
B+2C=3......(iv)
and 6((B+C))+3B+2C=16B6C+3B+2C=1
3B4C=1..........(v)
On solving Eqs. (iv)and (v), we get C=4
On putting the value of C in Eq. (iv), we get B+2×4=3B=5
Putting the value of B and C in Eq. (i), we get A +(-5)+4 =0 A=1A=1,B=5,C=4
Now, 3x1(x1)(x2)(x3)dx=(A(x1)+B(x2)+C(x3))dx
=1(x1)dx+(5)(x2)dx+4(x3)dx=log|x1|5log|x2|+4log|x3|+C(1xdx=logx)

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image