# Evaluate the integral. integrate from -1 to 4 of (1)/((x+7)(x^2+4))dx

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`(1)/((x+7)(x^2+4)) = A/(x+7)+(Bx+C)/(x^2+4)`

`1 = A(x^2+4)+(Bx+C)(x+7)`

By compairing components;

x^2-----> 0 = A+B -----(1)

x -------> 0 = 7B+C -----(2)

constant --> 1 = 4A+7C ----(3)

By solving (1),(2) and (3) youl will get;

A = 1/53

B = -1/53

C = 7/53

`int^4_(-1) (1)/((x+7)(x^2+4))dx`

`= int^4_(-1)[(1/53)/(x+7)-((1/53)(x-7))/(x^2+4)]dx`

`= 1/53int^4_(-1) 1/(x+7)dx - 1/53int^4_(-1) (x-7)/(x^2+4)dx`

`= 1/53int^4_(-1) 1/(x+7)dx - 1/53int^4_(-1)[(1/2)(2x)/(x^2+4)-7/(x^2+4)]dx`

`= 1/53int^4_(-1) 1/(x+7)dx - 1/53int^4_(-1) (1/2)(2x)/(x^2+4)+7/53int^4_(-1) 1/(x^2+2^2)dx`

`= [1/53ln(x+7)-1/53ln(x^2+4)+7/53*1/2tan^(-1)(x/2)]^4_(-1)`

`= 1/53(ln11-ln6)-1/53(ln20-ln5)+7/106(tan^(-1)2-tan^(-1)(-1/2))`

`= 1/53ln(11/6)-1/53ln(20/5)+7/106(0.352pi+0.148pi)`

`= 1/53ln(11/24)+7/106*0.5`

`= 1/53ln(11/24)+7/212`

`= 0.0183`

`int^4_(-1) (1)/((x+7)(x^2+4))dx = 0.0183`

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