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

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use partial fraction decomposition that helps you to evaluate two simpler integrals instead of one much harder integral such that:

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

`1 = Ax^2 + 4A + Bx^2 + Cx + 7Bx + 7C`

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

Equating the coefficients of like powers yields:

`A+B = 0`

`C + 7B = 0 => C/7 = -B`

`4A + 7C = 1 => A = (1-7C)/4`

`(1-7C)/4 = -B => C/7 = (1-7C)/4 => 7(1-7C) = 4C`

`7 - 49C = 4C => 7 = 53C => C = 7/53 => B = -1/53 => A = 1/53`

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

Integrating both sides yields:

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

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

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

You should solve the integral `int -x/(x^2+4)dx ` using the following substitution such that:

`x^2 + 4 = t => 2xdx = dt => xdx = (dt)/2`

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

`int 7/(x^2+4)dx = 7tan^(-1) (x/2) + c`

`int 1/((x+7)(x^2+4)) dx = (1/53)(ln|x+7| - (1/2) ln(x^2+4) + 7tan^(-1) (x/2) + c)`

You may evaluate the definite integral such that:

`int_(-1)^4 1/((x+7)(x^2+4)) dx = (1/53)(ln 11- (1/2) ln(20) + 7tan^(-1) (2) - ln 6+ (1/2) ln 5 + 7tan^(-1) (1/2))`

`int_(-1)^4 1/((x+7)(x^2+4)) dx = (1/53)(ln (11/6) + ln sqrt(1/4) + 7(tan^(-1) (2) + tan^(-1) (1/2)))`

Hence, evaluating the given definite integral under the givn conditions yields `int_(-1)^4 1/((x+7)(x^2+4)) dx = (1/53)(ln (11/6) + ln sqrt(1/4) + 7(tan^(-1) (2) + tan^(-1) (1/2))).`