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$\int_1^3 x^n \sqrt{x^2-1} d x=6$, then $n=$
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Verified Answer
The correct answer is:
$3$
Let $I=\int_1^3 x \sqrt[n]{x^2-1} d x=6$
$=\frac{1}{2} \int_1^3 2 x\left(x^2-1\right)^{\frac{1}{n}} d x=6$
$\Rightarrow\left[\frac{\left(n^2-1\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}\right]_1^3=12$
$\Rightarrow \quad \frac{(8)^{\frac{1}{n}+1}}{\frac{1}{n}+1}=12$
$\Rightarrow \quad \frac{(8)^{\frac{1}{n}}}{(n+1)}(n)=\frac{3}{2}$
Now, on putting $n=3$,
L H S $=$ R. H S
$\therefore n=3$
$=\frac{1}{2} \int_1^3 2 x\left(x^2-1\right)^{\frac{1}{n}} d x=6$
$\Rightarrow\left[\frac{\left(n^2-1\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}\right]_1^3=12$
$\Rightarrow \quad \frac{(8)^{\frac{1}{n}+1}}{\frac{1}{n}+1}=12$
$\Rightarrow \quad \frac{(8)^{\frac{1}{n}}}{(n+1)}(n)=\frac{3}{2}$
Now, on putting $n=3$,
L H S $=$ R. H S
$\therefore n=3$
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