Question

If \(\displaystyle \int \frac{x\,dx}{\sqrt[15]{(1+x^2)^{12}(2+x^2)^{18}}}=\alpha\left(\frac{1+x^2}{2+x^2}\right)^{1/n}+C\), then \(\dfrac{n}{\alpha}=\)

• A. \(6\)
• B. \(4\)
• C. \(2\)
• D. \(8\)

Hint:

When the integrand contains powers of \(1+x^2\) and \(2+x^2\), try the substitution \[ u=\frac{1+x^2}{2+x^2}. \] This substitution is useful because its derivative contains \[ \frac{x\,dx}{(2+x^2)^2}. \]

Answer 1

The Correct Option is C

Step 1: Simplify the denominator.
Given integral is \[ \int \frac{x\,dx}{\sqrt[15]{(1+x^2)^{12}(2+x^2)^{18}}} \] Now, \[ \sqrt[15]{(1+x^2)^{12}(2+x^2)^{18}} = (1+x^2)^{12/15}(2+x^2)^{18/15} \] \[ =(1+x^2)^{4/5}(2+x^2)^{6/5} \] So the integral becomes \[ \int \frac{x\,dx}{(1+x^2)^{4/5}(2+x^2)^{6/5}} \]

Step 2: Use substitution.
Let \[ u=\frac{1+x^2}{2+x^2} \] Differentiating, \[ \frac{du}{dx} = \frac{(2+x^2)(2x)-(1+x^2)(2x)}{(2+x^2)^2} \] \[ \frac{du}{dx} = \frac{2x[(2+x^2)-(1+x^2)]}{(2+x^2)^2} \] \[ \frac{du}{dx} = \frac{2x}{(2+x^2)^2} \] Hence, \[ du=\frac{2x}{(2+x^2)^2}\,dx \] So, \[ \frac{x\,dx}{(2+x^2)^2}=\frac{du}{2} \]

Step 3: Convert the integral in terms of \(u\).
Since \[ u=\frac{1+x^2}{2+x^2}, \] we can write \[ 1+x^2=u(2+x^2) \] Therefore, \[ (1+x^2)^{4/5}=(u(2+x^2))^{4/5} \] \[ =(u)^{4/5}(2+x^2)^{4/5} \] Thus, \[ (1+x^2)^{4/5}(2+x^2)^{6/5} = u^{4/5}(2+x^2)^{4/5+6/5} \] \[ =u^{4/5}(2+x^2)^2 \] Hence, \[ \int \frac{x\,dx}{(1+x^2)^{4/5}(2+x^2)^{6/5}} = \int \frac{x\,dx}{u^{4/5}(2+x^2)^2} \] \[ =\frac12\int u^{-4/5}\,du \]

Step 4: Integrate.
\[ \frac12\int u^{-4/5}\,du = \frac12\cdot \frac{u^{1/5}}{1/5}+C \] \[ =\frac{5}{2}u^{1/5}+C \] Substituting back, \[ =\frac{5}{2}\left(\frac{1+x^2}{2+x^2}\right)^{1/5}+C \]

Step 5: Compare with the given form.
Given, \[ \int \frac{x\,dx}{\sqrt[15]{(1+x^2)^{12}(2+x^2)^{18}}} = \alpha\left(\frac{1+x^2}{2+x^2}\right)^{1/n}+C \] Comparing, \[ \alpha=\frac52 \] and \[ n=5 \] Therefore, \[ \frac{n}{\alpha} = \frac{5}{5/2} \] \[ =2 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{2} \]