# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{1}{x\left(x^{n}+1\right)} d x$

Solution:

$\frac{1}{x\left(x^{n}+1\right)}$

Multiply numerator and denominator by $x^{n-1}$, we get

$\int \frac{1}{x\left(x^{n}+1\right)} d x \Rightarrow \int \frac{x^{n-1}}{x\left(x^{n}+1\right) x^{n-1}} d x \Rightarrow \int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x$

Let $x^{n}=t \Rightarrow n x^{n-1} d x=d t$

So the above equation becomes,

$\int \frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)} \mathrm{dx} \Rightarrow \frac{1}{\mathrm{n}} \int \frac{1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$

The denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1} \ldots \ldots(i)$

$\Rightarrow \frac{1}{t(t+1)}=\frac{A(t+1)+B t}{t(t+1)}$

$\Rightarrow 1=\mathrm{A}(\mathrm{t}+1)+\mathrm{Bt} \ldots \ldots$ (ii)

Put $t=0$ in above equations we get

$1=A(0+1)+B(0)$

$\Rightarrow A=1$

Now put $t=-1$ in equation (ii) we get

$1=A(-1+1)+B(-1)$

$\Rightarrow B=-1$

We put the values of $A$ and $B$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int \frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)} \mathrm{dx} \Rightarrow \frac{1}{\mathrm{n}} \int \frac{1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$

$\Rightarrow \frac{1}{n} \int\left[\frac{A}{t}+\frac{B}{t+1}\right] d t$

$\Rightarrow \frac{1}{n} \int\left[\frac{1}{t}+\frac{-1}{t+1}\right] d t$

Split up the integral,

$\Rightarrow \frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right]$

Let substitute

$\mathrm{u}=\mathrm{t}+1 \Rightarrow \mathrm{du}=\mathrm{dt}$, so the above equation becomes,

$\Rightarrow \frac{1}{\mathrm{n}}\left[\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{u}} \mathrm{du}\right]$

Substituting back the values of $u$, we get

$\Rightarrow \frac{1}{\mathrm{n}}[\log |\mathrm{t}|-\log (|\mathrm{t}+1|)]+\mathrm{C}$

Substituting back the values of $t$, we get

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

Applying the logarithm rules, we get

$\Rightarrow \frac{1}{n}\left[\log \left|\frac{x^{n}}{x^{n}+1}\right|\right]+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n}\left[\log \left|\frac{x^{n}}{x^{n}+1}\right|\right]+C$