First slide

Methods of integration

Question

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

Moderate

A

$\log |\frac{x^{n} + 1}{x^{n}}| + C$
B

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

$\frac{1}{n} \log |\frac{x^{n}}{x^{n} + 1}| + C$
D

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

Solution

Here, partial fractions of given integrand cannot find easily, so we convert it into easy form for partial fractions by multiply x^n-1 in both numerator and denominator and then put xⁿ =t. Now, apply partial fractions method and then integrate.
let $I = \int \frac{1}{x (x^{n} + 1)} dx = \int \frac{x^{n - 1}}{x^{n} (x^{n} + 1)} dx$
put $x^{n} = t \Rightarrow {nx}^{n - 1} dx = dt \Rightarrow x^{n - 1} dx = \frac{1}{n} dt$
$∴ I = \int \frac{x^{n - 1}}{x^{n} (x^{n} + 1)} dx = \frac{1}{n} \int \frac{1}{t (t + 1)} dt - - - - (i)$
$Now, \frac{1}{t (t + 1)} = \frac{A}{t} + \frac{B}{(t + 1)}$
$\Rightarrow 1 = A (1 + t) + Bt - - - - - (ii)$
On comparing, we get
$\begin{array}{l} A + B = 0, A = 1 \\ \Rightarrow B = - 1 \\ ∴ I = \frac{1}{n} \int [\frac{1}{t} - \frac{1}{(t + 1)}] \cdot dt = \frac{1}{n} [\log | t | - \log | t + 1 |] + C \\ = \frac{1}{n} \log |\frac{t}{t + 1}| + C = \frac{1}{n} \log |\frac{x^{n}}{x^{n} + 1}| + C \end{array}$

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