Integral of (x^23)/(x^32+1)?

Mar 18, 2018

$\frac{1}{4 \sqrt{2}} a r c \left(\frac{{x}^{16} - 1}{\sqrt{2} \cdot {x}^{8}}\right)$
$+ \frac{1}{8 \sqrt{2}} \ln | \frac{{x}^{16} - \sqrt{2} \cdot {x}^{8} + 1}{{x}^{16} + \sqrt{2} \cdot {x}^{8} + 1} | + C$.

Explanation:

Let, $I = \int {x}^{23} / \left({x}^{32} + 1\right) \mathrm{dx}$.

$\frac{d}{\mathrm{dx}} \left({x}^{24}\right) = 24 {x}^{23}$ suggests that the substitution ${x}^{24} = y$ may work.

${x}^{24} = y \Rightarrow 24 {x}^{23} \mathrm{dx} = \mathrm{dy}$.

$\therefore I = \frac{1}{24} \int \frac{1}{{\left({y}^{\frac{1}{24}}\right)}^{32} + 1} \mathrm{dy} = \frac{1}{24} \int \frac{1}{{y}^{\frac{4}{3}} + 1} \mathrm{dy}$.

To get rid of cube root, we let,

${y}^{\frac{1}{3}} = t \Rightarrow \frac{1}{3} \cdot {y}^{- \frac{2}{3}} \mathrm{dy} = \mathrm{dt} \Rightarrow \mathrm{dy} = 3 {y}^{\frac{2}{3}} \mathrm{dt} = 3 {t}^{2} \mathrm{dt}$.

$\therefore I = \frac{3}{24} \int {t}^{2} / \left({t}^{4} + 1\right) \mathrm{dt} = \frac{1}{8} \int {t}^{2} / \left({t}^{4} + 1\right) \mathrm{dt}$,

$= \frac{1}{4} \int \frac{2 {t}^{2}}{{t}^{4} + 1} \mathrm{dt} = \frac{1}{4} \int \frac{\left({t}^{2} + 1\right) + \left({t}^{2} - 1\right)}{{t}^{4} + 1} \mathrm{dt}$.

$\therefore 4 I = \int \frac{{t}^{2} + 1}{{t}^{4} + 1} \mathrm{dt} + \int \frac{{t}^{2} - 1}{{t}^{4} + 1} \mathrm{dt} = {I}_{1} + {I}_{2} , s a y$.

I_1=int(t^2+1)/(t^4+1)dt=int{t^2(1+1/t^2)}/{t^2(t^2+1/t^2)dt.

$\therefore {I}_{1} = \int \frac{{t}^{2} + \frac{1}{t} ^ 2}{{t}^{2} + \frac{1}{t} ^ 2} \mathrm{dt} = \int \frac{\frac{d}{\mathrm{dt}} \left(t - \frac{1}{t}\right)}{{\left(t - \frac{1}{t}\right)}^{2} + 2} \mathrm{dt}$,

$= \int \frac{\mathrm{du}}{{u}^{2} + 2} \ldots \ldots \ldots \ldots \left[u = t - \frac{1}{t}\right]$,

$= \frac{1}{\sqrt{2}} a r c \frac{u}{\sqrt{2}}$,

$= \frac{1}{\sqrt{2}} a r c \left(\frac{{t}^{2} - 1}{\sqrt{2} t}\right)$,

$= \frac{1}{\sqrt{2}} a r c \left(\frac{{y}^{\frac{2}{3}} - 1}{\sqrt{2} \cdot {u}^{\frac{1}{3}}}\right)$,

$= \frac{1}{\sqrt{2}} a r c \left(\frac{{x}^{16} - 1}{\sqrt{2} \cdot {x}^{8}}\right)$.

Further, ${I}_{2} = \int \frac{{t}^{2} - 1}{{t}^{4} + 1} \mathrm{dt}$

$= \int \frac{{t}^{2} \left(1 - \frac{1}{t} ^ 2\right)}{{t}^{2} \left({t}^{2} + \frac{1}{t} ^ 2\right)} \mathrm{dt}$,

$= \int \frac{1 - \frac{1}{t} ^ 2}{{t}^{2} + \frac{1}{t} ^ 2} \mathrm{dt}$,

$= \int \frac{\frac{d}{\mathrm{dt}} \left(t + \frac{1}{t}\right)}{{\left(t + \frac{1}{t}\right)}^{2} - 2} \mathrm{dt}$,

$= \int \frac{\mathrm{dv}}{{v}^{2} - 2} \mathrm{dv} \ldots \ldots \ldots \ldots . \left[v = t + \frac{1}{t}\right]$,

$= \frac{1}{2 \sqrt{2}} \ln | \frac{v - \sqrt{2}}{v + \sqrt{2}} |$,

$= \frac{1}{2 \sqrt{2}} \ln | \frac{{t}^{2} - \sqrt{2} t + 1}{{t}^{2} + \sqrt{2} t + 1} |$,

$= \frac{1}{2 \sqrt{2}} \ln | \frac{{y}^{\frac{2}{3}} - \sqrt{2} \cdot {y}^{\frac{1}{3}} + 1}{{y}^{\frac{2}{3}} + \sqrt{2} \cdot {y}^{\frac{1}{3}} + 1} |$,

$= \frac{1}{2 \sqrt{2}} \ln | \frac{{x}^{16} - \sqrt{2} \cdot {x}^{8} + 1}{{x}^{16} + \sqrt{2} \cdot {x}^{8} + 1} |$.

Finally, we get,

$I = \frac{1}{4 \sqrt{2}} a r c \left(\frac{{x}^{16} - 1}{\sqrt{2} \cdot {x}^{8}}\right)$
$+ \frac{1}{8 \sqrt{2}} \ln | \frac{{x}^{16} - \sqrt{2} \cdot {x}^{8} + 1}{{x}^{16} + \sqrt{2} \cdot {x}^{8} + 1} | + C$.

Enjoy Maths.!