Question

Evaluate \[ \int \frac{dx}{(x-3)^{4/5}(x+1)^{6/5}} \]

• A. \(\frac{5}{4}\sqrt[5]{\frac{x-3}{x+1}}+C\)
• B. \(\frac{5}{4}\left(\frac{x+1}{x-3}\right)^{1/5}+C\)
• C. \(\frac{1}{5}\left(\frac{x-3}{x+1}\right)^{1/5}+C\)
• D. \(\frac{5}{4}\left(\frac{x-3}{x+4}\right)^{4/5}+C\)

Hint:

When the integrand contains powers of two linear factors, try substituting a fractional power of their ratio.

Answer 1

The Correct Option is A

Step 1: Choose a suitable substitution.
Let \[ u=\left(\frac{x-3}{x+1}\right)^{1/5} \] Then, \[ u=\sqrt[5]{\frac{x-3}{x+1}} \]

Step 2: Differentiate \(u\).
Now, \[ u=\left(\frac{x-3}{x+1}\right)^{1/5} \] Differentiate with respect to \(x\): \[ \frac{du}{dx} = \frac{1}{5}\left(\frac{x-3}{x+1}\right)^{-4/5} \cdot \frac{(x+1)-(x-3)}{(x+1)^2} \] \[ \frac{du}{dx} = \frac{1}{5}\left(\frac{x-3}{x+1}\right)^{-4/5} \cdot \frac{4}{(x+1)^2} \] \[ \frac{du}{dx} = \frac{4}{5} \left(\frac{x+1}{x-3}\right)^{4/5} \cdot \frac{1}{(x+1)^2} \] \[ \frac{du}{dx} = \frac{4}{5} \cdot \frac{1}{(x-3)^{4/5}(x+1)^{6/5}} \]

Step 3: Relate this with the given integral.
Thus, \[ \frac{dx}{(x-3)^{4/5}(x+1)^{6/5}} = \frac{5}{4}du \] Therefore, \[ \int \frac{dx}{(x-3)^{4/5}(x+1)^{6/5}} = \frac{5}{4}\int du \] \[ = \frac{5}{4}u+C \]

Step 4: Substitute back the value of \(u\).
Hence, \[ \frac{5}{4}u+C = \frac{5}{4}\left(\frac{x-3}{x+1}\right)^{1/5}+C \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{5}{4}\sqrt[5]{\frac{x-3}{x+1}}+C} \]