We read your photo as:

Solve in english

English

ScanSolve AI Answer

Answers

2 parts
a

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

b

ln⁡(32)\ln \left(\frac{3}{2}\right)

Step-by-step solution

10 steps

)

Part a· 4 steps

Express f(x)f(x) in partial fractions where f(x)=2(x−1)(x+1)f(x) = \frac{2}{(x-1)(x+1)}.

(Assuming the 'b' in the denominator is a typo and the function is f(x)=2(x−1)(x+1)f(x) = \frac{2}{(x-1)(x+1)} to align with the subsequent integral.)

  1. Set up the partial fraction decomposition.

    We want to find constants AA and BB such that:

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

  2. Combine the terms on the right-hand side.

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

    Equating the numerators:

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

  3. Solve for AA and BB.

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

    2=A(1+1)+B(1−1)2 = A(1+1) + B(1-1) 2=2A+02 = 2A + 0 A=1A = 1

    To find BB, let x=−1x=-1:

    2=A(−1+1)+B(−1−1)2 = A(-1+1) + B(-1-1) 2=0−2B2 = 0 - 2B B=−1B = -1

  4. Write the partial fraction decomposition.

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

Part b

ln⁡(32)\ln \left(\frac{3}{2}\right)

Hence using the substitution x=t2−2x=t^2-2, or otherwise, show that ∫271(x+1)x+2dx=ln⁡(32)\int_2^7 \frac{1}{(x+1)\sqrt{x+2}} dx = \ln \left(\frac{3}{2}\right).

  1. Apply the substitution x=t2−2x=t^2-2.

    If x=t2−2x = t^2-2, then dx=2t dtdx = 2t \, dt.

  2. Change the limits of integration.

    When x=2x=2:

    2=t2−2  ⟹  t2=4  ⟹  t=2(since t≥0 for x+2)2 = t^2 - 2 \implies t^2 = 4 \implies t = 2 \quad (\text{since } t \ge 0 \text{ for } \sqrt{x+2})

    When x=7x=7:

    7=t2−2  ⟹  t2=9  ⟹  t=3(since t≥0)7 = t^2 - 2 \implies t^2 = 9 \implies t = 3 \quad (\text{since } t \ge 0)

  3. Substitute into the integral.

    The term x+2\sqrt{x+2} becomes:

    x+2=(t2−2)+2=t2=t\sqrt{x+2} = \sqrt{(t^2-2)+2} = \sqrt{t^2} = t

    The term (x+1)(x+1) becomes:

    x+1=(t2−2)+1=t2−1x+1 = (t^2-2)+1 = t^2-1

    Substitute these into the integral:

    ∫271(x+1)x+2dx=∫231(t2−1)t(2t dt)\int_2^7 \frac{1}{(x+1)\sqrt{x+2}} dx = \int_2^3 \frac{1}{(t^2-1)t} (2t \, dt) =∫232t(t2−1)tdt= \int_2^3 \frac{2t}{(t^2-1)t} dt =∫232t2−1dt= \int_2^3 \frac{2}{t^2-1} dt =∫232(t−1)(t+1)dt= \int_2^3 \frac{2}{(t-1)(t+1)} dt

  4. Use the partial fraction decomposition from part (a).

    From part (a), we know that 2(t−1)(t+1)=1t−1−1t+1\frac{2}{(t-1)(t+1)} = \frac{1}{t-1} - \frac{1}{t+1}.
    So the integral becomes:

    ∫23(1t−1−1t+1)dt\int_2^3 \left(\frac{1}{t-1} - \frac{1}{t+1}\right) dt

  5. Integrate the partial fractions.

    [ln⁡∣t−1∣−ln⁡∣t+1∣]23\left[\ln|t-1| - \ln|t+1|\right]_2^3

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

    [ln⁡∣t−1t+1∣]23\left[\ln\left|\frac{t-1}{t+1}\right|\right]_2^3

  6. Evaluate the definite integral.

    Substitute the upper limit t=3t=3:

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

    Substitute the lower limit t=2t=2:

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

    Subtract the lower limit value from the upper limit value:

    ln⁡(12)−ln⁡(13)=ln⁡(1/21/3)\ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right) = \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)

    Thus, we have shown that ∫271(x+1)x+2dx=ln⁡(32)\int_2^7 \frac{1}{(x+1)\sqrt{x+2}} dx = \boxed{\ln \left(\frac{3}{2}\right)}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More English Questions

QSolve in English languageQSolve in english languagQSolve it in English
See all English questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.