∫dxxxn+1 is equal to

# $\int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}+1\right)}$ is equal to

1. A

$\mathrm{log}\left|\frac{{\mathrm{x}}^{\mathrm{n}}+1}{{\mathrm{x}}^{\mathrm{n}}}\right|+\mathrm{C}$

2. B

$-\mathrm{log}\left|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}+1}\right|+\mathrm{C}$

3. C

$\frac{1}{\mathrm{n}}\mathrm{log}\left|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}+1}\right|+\mathrm{C}$

4. D

$\mathrm{nlng}\left|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}+1}\right|+\mathrm{C}$

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### Solution:

Here, partial fractions of given integrand cannot find easily, so we convert it into easy form for partial fractions by multiply xn-1 in both numerator and denominator and then put xn =t. Now, apply partial fractions method and then integrate.

let  $\mathrm{I}=\int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}+1\right)}\mathrm{dx}=\int \frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{x}}^{\mathrm{n}}+1\right)}\mathrm{dx}$

put ${\mathrm{x}}^{\mathrm{n}}=\mathrm{t}⇒{\mathrm{nx}}^{\mathrm{n}-1}\mathrm{dx}=\mathrm{dt}⇒{\mathrm{x}}^{\mathrm{n}-1}\mathrm{dx}=\frac{1}{\mathrm{n}}\mathrm{dt}$

On comparing, we get

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