Question

\[ \int \frac{\mathrm{d} x}{(x+\mathrm{a})^{\frac{9}{7}}(x-\mathrm{b})^{\frac{5}{7}}}= \]

• A. \(\frac{7}{\mathrm{a}+\mathrm{b}}(\frac{x-\mathrm{b}}{x+\mathrm{a}})^{\frac{9}{7}}+\mathrm{c}\), where c is the constant of integration.
• B. \(\frac{7}{a+b}(\frac{x-b}{x+a})^{\frac{5}{7}}+c\), where \(c\) is the constant of integration.
• C. \(\frac{7}{2(a+b)}(\frac{x-b}{x+a})^{\frac{2}{7}}+c\), where \(c\) is the constant of integration.
• D. \(\frac{7}{a+b}(\frac{x-b}{x+a})^{\frac{1}{7}}+c\), where \(c\) is the constant of integration.

Hint:

The powers add up to 2, so factor out (x+a) squared and substitute t = (x-b)/(x+a).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The powers \(\dfrac97\) and \(\dfrac57\) add up to \(2\). So write the integrand as \(\dfrac{1}{(x+a)^2}\left(\dfrac{x-b}{x+a}\right)^{-5/7}\).

Step 2: Substitute
Put \(t=\dfrac{x-b}{x+a}\). Then \(dt=\dfrac{(x+a)-(x-b)}{(x+a)^2}dx=\dfrac{a+b}{(x+a)^2}dx\).

Step 3: Calculation
\(\int=\dfrac{1}{a+b}\int t^{-5/7}dt=\dfrac{1}{a+b}\cdot\dfrac{t^{2/7}}{2/7}=\dfrac{7}{2(a+b)}t^{2/7}+c\)

Step 4: Compare the options
This matches option (C).

Final Answer:
The integral equals 7/(2(a+b)) times ((x-b)/(x+a)) to the power 2/7, plus c. \[ \boxed{\text{(C) }\dfrac{7}{2(a+b)}\left(\dfrac{x-b}{x+a}\right)^{\frac27}+c} \]