Question

If \(\int \frac{dx}{x^{7/2}(x^4+1)^{3/8}} = m(\frac{x^4+1}{x^4})^n+c\), where \(c\) is a constant of integration, then the value of \(\frac{n}{m}\) is...

• A. \(-\frac{1}{16}\)
• B. \(-\frac{25}{16}\)
• C. \(\frac{25}{4}\)
• D. \(-\frac{25}{4}\)

Hint:

Factor x^4 out of the bracket and substitute t = 1 + x^-4.

Answer 1

The Correct Option is B

Step 1: Rewrite the Integrand:
\((x^4+1)^{3/8}=x^{3/2}(1+x^{-4})^{3/8}\). So the denominator becomes \(x^{7/2}\cdot x^{3/2}(1+x^{-4})^{3/8}=x^5(1+x^{-4})^{3/8}\).
\[ I=\int x^{-5}(1+x^{-4})^{-3/8}dx \]

Step 2: Substitute:
Let \(t=1+x^{-4}\). Then \(dt=-4x^{-5}dx\), so \(x^{-5}dx=-\tfrac14dt\).

Step 3: Integrate:
\[ I=-\frac14\int t^{-3/8}dt=-\frac14\cdot\frac{t^{5/8}}{5/8}=-\frac25t^{5/8}+c \]

Step 4: Compare With the Given Form:
\(\dfrac{x^4+1}{x^4}=t\), so \(I=m\,t^n+c\) with \(m=-\dfrac25\) and \(n=\dfrac58\).
\[ \frac nm=\frac{5/8}{-2/5}=-\frac{25}{16} \]

Final Answer:
The ratio is \(-\dfrac{25}{16}\), option (B). \[ \boxed{\text{(B) } -\frac{25}{16}} \]