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Question
Evaluate: `∫8/((x+2)(x^2+4))dx`
Solution
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`
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