Question

If \[ \int_1^3 x\sqrt[n]{x^2-1}\,dx=6, \] then \(n=\)

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

For integrals involving expressions like \(x^2-a^2\), the substitution \[ u=x^2-a^2 \] usually simplifies the integral immediately.

Answer 1

The Correct Option is B

Step 1: Use substitution.
Given, \[ \int_1^3 x\sqrt[n]{x^2-1}\,dx=6 \] Let \[ u=x^2-1 \] Then, \[ du=2x\,dx \] or \[ x\,dx=\frac{du}{2} \]

Step 2: Change the limits.
When \[ x=1, \] we get \[ u=1^2-1=0 \] When \[ x=3, \] we get \[ u=3^2-1=8 \] Therefore, \[ \int_1^3 x\sqrt[n]{x^2-1}\,dx = \frac12\int_0^8 u^{1/n}\,du \]

Step 3: Integrate.
\[ \frac12\int_0^8 u^{1/n}\,du = \frac12\left[ \frac{u^{\frac{n+1}{n}}}{\frac{n+1}{n}} \right]_0^8 \] \[ = \frac12\cdot \frac{n}{n+1} \left(8^{\frac{n+1}{n}}\right) \] Given that the value is \(6\), \[ \frac12\cdot \frac{n}{n+1}\cdot 8^{\frac{n+1}{n}}=6 \]

Step 4: Check the options.
For \[ n=3, \] \[ \frac12\cdot \frac{3}{4}\cdot 8^{4/3} \] Since \[ 8^{4/3}=(\sqrt[3]{8})^4=2^4=16, \] we get \[ \frac12\cdot \frac34\cdot 16 = \frac38\cdot 16 =6 \] Hence, \[ n=3 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{3} \]