Calculate the indefinite integral of 1/(x^2-7) .

Expert Answers

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Let f(x) = 1/(x^2 -7)

Then.

F(x) = intg f(x)

         = intg 1/(x^2 -7)

But, we know that:

x^2 - 7 = (x-sqrt7)(x+sqrt7)

==> F(x) = intg 1/(x-sqrt7)(x+sqrt7)

Let us rewrite:

A/(x-sqrt7) + B/(x+sqrt7) = 1/(x-sqrt7)(x+sqrt7)

==> A(x+sqrt7) + B(x-sqrt7) = 1

==> Ax + Asqrt7 + Bx - Bsqrt7 = 1

==> (A+B)x + (A-B)sqrt7 = 1

==> A+B = 0==> A = -B

==> (A-B)sqrt7 = 1

==> (-2B)sqrt7 = 1

==> B = -1/2sqrt7

==> A = 1/2sqrt7

==> F(x) = intg 1/2sqrt7(x-sqrt7) - 1/2sqrt7(x+sqrt7)

               = (1/2sqrt7)*log (x-sqrt7) - (1/2sqrt7)*log (x+sqrt7)

              = (1/2sqrt7)*[log (x-sqrt7) - log (x+sqrt7)]

                = (1/2sqrt7)*log [(x-sqrt7)/(x+sqrt7)]

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