Question

The value of $\displaystyle \int \frac{d x}{x \left(\right. x^{n} + 1 \left.\right)}$ is equal to

• A. $\frac{1}{n} \left(log\right)_{e} \left(\right. \frac{x^{n}}{x^{n} + 1} \left.\right) + c$
• B. $\frac{1}{n} \left(log\right)_{e} \left(\right. \frac{x^{n} + 1}{x^{n}} \left.\right) + c$
• C. $\left(log\right)_{e} \left(\right. \frac{x^{n}}{x^{n} + 1} \left.\right) + c$
• D. None of these

Answer 1

Answer: (A)

$\text{I=} \displaystyle \int \frac{d x}{\left(\right. 1 + \frac{1}{x^{n}} \left.\right) \times x^{n + 1}}$
put $1 + \frac{1}{x^{n}} = t$
$- \frac{n}{x^{n + 1}} . d x = d t$
$I = \displaystyle \int \frac{- \frac{d t}{n}}{t} = - \frac{1}{n} ln \left|\right. t \left|\right. + c$
$I = - \frac{1}{n} ln \left(\right. 1 + \frac{1}{x^{n}} \left.\right) + c$
$I = \frac{1}{n} ln \left(\right. \frac{x^{n}}{x^{n} + 1} \left.\right) + c$