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Given f(x) = (2)/((x-1)(x+1)), we write it in the form:

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2 parts
1

1x−1−1x+1\frac{1}{x-1} - \frac{1}{x+1}

2

ln⁡32\ln\frac{3}{2}

Graph of f(x) = 2/((x − 1)(x + 1))

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Key points of f(x) = 2/((x − 1)(x + 1))
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Step-by-step solution

10 steps

Here is the solution for both parts of the problem.

Part 1· 4 steps

Express f(x)f(x) in partial fractions.

  1. Set up the partial fraction decomposition.

    Given f(x)=2(x−1)(x+1)f(x) = \frac{2}{(x-1)(x+1)}, we write it in the form:

    2(x−1)(x+1)=Ax−1+Bx+1\frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}

  2. Clear the denominators.

    Multiply both sides by (x−1)(x+1)(x-1)(x+1):

    2=A(x+1)+B(x−1)2 = A(x+1) + B(x-1)

  3. Solve for the constants AA and BB.

    To find AA, set x=1x=1:

    2=A(1+1)+B(1−1)2 = A(1+1) + B(1-1) 2=2A2 = 2A A=1A = 1

    To find BB, set x=−1x=-1:

    2=A(−1+1)+B(−1−1)2 = A(-1+1) + B(-1-1) 2=−2B2 = -2B B=−1B = -1

  4. Write the final partial fraction expression.

    Substitute the values of AA and BB back into the decomposition:

    f(x)=1x−1−1x+1f(x) = \frac{1}{x-1} - \frac{1}{x+1}

    The partial fraction decomposition of f(x)f(x) is 1x−1−1x+1\boxed{\frac{1}{x-1} - \frac{1}{x+1}}.

Part 2

ln⁡32\ln\frac{3}{2}

Evaluate the definite integral.

  1. Apply the substitution.

    Let u=x+2u = \sqrt{x+2}.
    Then u2=x+2u^2 = x+2, which implies x=u2−2x = u^2 - 2.
    From this, x+1=(u2−2)+1=u2−1x+1 = (u^2 - 2) + 1 = u^2 - 1.
    Differentiating u2=x+2u^2 = x+2 with respect to xx, we get 2u du=dx2u\,du = dx.

  2. Change the limits of integration.

    When x=2x=2, u=2+2=4=2u = \sqrt{2+2} = \sqrt{4} = 2.
    When x=7x=7, u=7+2=9=3u = \sqrt{7+2} = \sqrt{9} = 3.

  3. Rewrite the integral in terms of uu.

    ∫271(x+1)x+2 dx=∫231(u2−1)u(2u du)\int_{2}^{7} \frac{1}{(x+1)\sqrt{x+2}}\,dx = \int_{2}^{3} \frac{1}{(u^2-1)u} (2u\,du) =∫232uu(u2−1) du= \int_{2}^{3} \frac{2u}{u(u^2-1)}\,du =∫232u2−1 du= \int_{2}^{3} \frac{2}{u^2-1}\,du

  4. Use the partial fraction decomposition from Part 1.

    From Part 1, we know that 2u2−1=1u−1−1u+1\frac{2}{u^2-1} = \frac{1}{u-1} - \frac{1}{u+1}.
    Substitute this into the integral:

    ∫23(1u−1−1u+1) du\int_{2}^{3} \left(\frac{1}{u-1} - \frac{1}{u+1}\right)\,du

  5. Integrate the expression.

    [ln⁡∣u−1∣−ln⁡∣u+1∣]23\left[\ln|u-1| - \ln|u+1|\right]_{2}^{3}

    Using the logarithm property ln⁡a−ln⁡b=ln⁡ab\ln a - \ln b = \ln \frac{a}{b}:

    =[ln⁡∣u−1u+1∣]23= \left[\ln\left|\frac{u-1}{u+1}\right|\right]_{2}^{3}

  6. Evaluate the definite integral using the limits.

    ln⁡∣3−13+1∣−ln⁡∣2−12+1∣\ln\left|\frac{3-1}{3+1}\right| - \ln\left|\frac{2-1}{2+1}\right| =ln⁡∣24∣−ln⁡∣13∣= \ln\left|\frac{2}{4}\right| - \ln\left|\frac{1}{3}\right| =ln⁡(12)−ln⁡(13)= \ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right)

    Again, using the logarithm property ln⁡a−ln⁡b=ln⁡ab\ln a - \ln b = \ln \frac{a}{b}:

    =ln⁡(1/21/3)= \ln\left(\frac{1/2}{1/3}\right) =ln⁡(12×31)= \ln\left(\frac{1}{2} \times \frac{3}{1}\right) =ln⁡(32)= \ln\left(\frac{3}{2}\right)

    The value of the definite integral is ln⁡32\boxed{\ln\frac{3}{2}}.

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