Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{dx}}{{x\left( {{x^n} + 1} \right)}}}$

Now $\cfrac{1}{{x\left( {{x^n} + 1} \right)}} = \cfrac{{{x^{n - 1}}}}{{x \cdot {x^{n - 1}}\left( {{x^n} + 1} \right)}} = \cfrac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}$

Put ${x^n} = t$ $\Rightarrow$ $n{x^{n - 1}}dx = dt$

$\therefore$ $\int {\cfrac{{dx}}{{x\left( {{x^n} + 1} \right)}}} = \cfrac{1}{n} \cdot \int {\cfrac{{n{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}} dx + \cfrac{1}{n}\int {\cfrac{{dt}}{{t\left( {t + 1} \right)}}}$

….(i)
We write, $\cfrac{1}{{t\left( {t + 1} \right)}} = \cfrac{A}{t} + \cfrac{B}{{t + 1}}$

$\Rightarrow$ $1 = A\left( {t + 1} \right) + Bt$

…(ii)
Putting $t = 0$ in (ii),

we get $1 = A\left( {0 + 1} \right)$ $\Rightarrow$ $A = 1$

Putting $t = - 1$ in (ii),

we get
$1 = B\left( { - 1} \right)$ $\Rightarrow$ $B = - 1$

$\therefore$ $I = \int {\cfrac{{dx}}{{x\left( {{x^n} + 1} \right)}} = \cfrac{1}{n}} \int {\left( {\cfrac{1}{t} - \cfrac{1}{{t + 1}}} \right)dt}$

$= \cfrac{1}{n}\left[ {\log \left| t \right| - \log \left| {t + 1} \right|} \right] + C = \cfrac{1}{n}\log \left| {\cfrac{t}{{t + 1}}} \right| + C = \cfrac{1}{n}\log \left| {\cfrac{{{x^n}}}{{{x^n} + 1}}} \right| + C$