Solve the integral `int 8/(16-x^) dx`

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Solve `int 8/(16-x^2)dx`

Factor the denominator and pull the `8` outside the integral.

`=8int 1/((4-x)(x+4))dx`

Preform partial fraction decomposition on `1/((4-x)(x+4))` .





Sine the left hand coefficients must be equal to the right hand side coefficients, `(A+B)` must be equal to zero to make `x` vanish and `4(A-B)` must equal `1` .






`1/8=A, -1/8=B`

Then the integral becomes:

`=8int (1/8)[1/(4-x) -1/(x+4)]dx`

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

Use u-substitution on the first integral.

`4-x=u` , and `du=-dx`

on the 2nd integral.

`x+4=v` , `dv=dx`

`=-int 1/(u)du-int 1/(v)dv`





Then finally,

`int 8/(16-x^2)dx=ln((x+4)/(4-x))+C`

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