# Question #0053d

Dec 7, 2017

$\frac{\sqrt{2}}{4} \cdot \arctan \left(\frac{{t}^{2} - 1}{t \sqrt{2}}\right) + \frac{\sqrt{2}}{8} L n \left(\frac{{t}^{2} - t \sqrt{2} + 1}{{t}^{2} + t \sqrt{2} + 1}\right) + C$

#### Explanation:

$\int {t}^{2} / \left({t}^{4} + 1\right) \cdot \mathrm{dt}$

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

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

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

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

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

=$\frac{\sqrt{2}}{4} \cdot \arctan \left(\frac{t - \frac{1}{t}}{\sqrt{2}}\right) + C$

=$\frac{\sqrt{2}}{4} \cdot \arctan \left(\frac{{t}^{2} - 1}{t \sqrt{2}}\right) + C$

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

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

=$\frac{\sqrt{2}}{8} L n \left(t + \frac{1}{t} - \sqrt{2}\right) - \frac{\sqrt{2}}{8} L n \left(t + \frac{1}{t} + \sqrt{2}\right)$

=$\frac{\sqrt{2}}{8} L n \left(\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}\right)$

=$\frac{\sqrt{2}}{8} L n \left(\frac{{t}^{2} - t \sqrt{2} + 1}{{t}^{2} + t \sqrt{2} + 1}\right)$

Thus,

$\int {t}^{2} / \left({t}^{4} + 1\right) \cdot \mathrm{dt}$

=$A + B$

=$\frac{\sqrt{2}}{4} \cdot \arctan \left(\frac{{t}^{2} - 1}{t \sqrt{2}}\right) + \frac{\sqrt{2}}{8} L n \left(\frac{{t}^{2} - t \sqrt{2} + 1}{{t}^{2} + t \sqrt{2} + 1}\right) + C$

Dec 7, 2017

$\frac{1}{2 \sqrt{2}} a r c \tan \left\{\frac{1}{\sqrt{2}} \left(t - \frac{1}{t}\right)\right\} + \frac{1}{4 \sqrt{2}} \ln | \frac{{t}^{2} - \sqrt{2} t + 1}{{t}^{2} + \sqrt{2} t + 1} | + C .$

#### Explanation:

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

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

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

$= \frac{1}{2} {I}_{1} + \frac{1}{2} {I}_{2.} \ldots \ldots \ldots . \left(\star\right) ,$ where,

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

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

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

Now, we substitute,

$\left(t - \frac{1}{t}\right) = u , \text{ so that, } \left(1 + \frac{1}{t} ^ 2\right) \mathrm{dt} = \mathrm{du} .$

Also, ${t}^{2} + \frac{1}{t} ^ 2 = {\left(t - \frac{1}{t}\right)}^{2} + 2.$

$\therefore {I}_{1} = \int \frac{1}{{u}^{2} + 2} \mathrm{du} = \frac{1}{\sqrt{2}} a r c \tan \left(\frac{u}{\sqrt{2}}\right) , i . e . ,$

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

$\Rightarrow {I}_{1} = \frac{1}{\sqrt{2}} a r c \tan \left(\frac{{t}^{2} - 1}{\sqrt{2} t}\right) \ldots \ldots . . \left({\star}_{1}\right) .$

Similarly, ${I}_{2} = \int \frac{1 - \frac{1}{t} ^ 2}{{t}^{2} + \frac{1}{t} ^ 2} \mathrm{dt} ,$ becomes, on using the

substitution $\left(t + \frac{1}{t}\right) = v , {I}_{2} = \int \frac{1}{{v}^{2} - 2} \mathrm{dv} ,$

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

$\Rightarrow {I}_{2} = \frac{1}{2 \sqrt{2}} \ln | \frac{{t}^{2} - \sqrt{2} t + 1}{{t}^{2} + \sqrt{2} t + 1} | \ldots \ldots . \left({\star}_{2}\right) .$

Altogether, from $\left(\star\right) , \left({\star}_{1}\right) \mathmr{and} \left({\star}_{2}\right) ,$ we have,

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

$+ \frac{1}{4 \sqrt{2}} \ln | \frac{{t}^{2} - \sqrt{2} t + 1}{{t}^{2} + \sqrt{2} t + 1} | + C .$

Enjoy Maths.!