Question

Evaluate: `∫8/((x+2)(x^2+4))dx` 

Answer 1

Let `I=∫8/((x+2)(x^2+4))dx` 

Let `8/((x+2)(x^2+4))=A/(x+2)+(Bx+C)/(x^2+4)` 

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

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

`8=(A+B)x^2+(2B+c)x+(4A+2C)`

Comparing the coefficients of x² , x and the constant term, we get
A + B = 0, 2B + C = 0 and 4A + 2C = 8
On solving these equations, we get
A = 1, B = –1, C = 2

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

`I=int[1/(x+2)+(-x+2)/(x^2+4)]dx`

`=int1/(x+2)dx-1/2int(2x)/(x^2+4)dx+2int1/(x^2+2^2)dx`

`=log|x+2|-1/2log|x^2+4|+tan^-1(x/2)+c`

`=log|(x+2)/sqrt(x^2+4)|+tan^-1(x/2)+c`