Navsari 396445, Gujarat, INDIA.
+91 94281 63811

Blog

Parin Sir > Mathematical Proofs > Indefinite Integration > Pattern - 26 of Indefinite Integration > Pattern - 26 of Indefinite Integration

Pattern - 26 of Indefinite Integration

February 19, 2021
Category: Pattern - 26 of Indefinite Integration

No Comments

⋆ Pattern – 26

$\int\frac{x^2+a}{x^4+bx^2+c\ }\ dx$

then divide numerator & denominator by & assume $x\pm\frac{k}{x}=t$

1) $\ I=\ \int\frac{x^2}{x^4+1}dx$

$=\frac{1}{2}\int\frac{2x^2}{x^4+1}\ dx$

$=\frac{1}{2}\ \int\frac{\left(x^2+1\right)+\left(x^2+1\right)}{x^4+1\ }dx$

$=\frac{1}{2}\ \int\frac{x^2+1}{x^4+1\ }\ dx+\frac{1}{2}\ \int\frac{x^2-1}{x^4+1}\ dx$

$\frac{1}{2}I_1+\frac{1}{2}I_2-----\left(1\right)$

Now,

$\ I_1=\ \int\frac{x^2+1}{x^4+1\ }\ dx$

$=\ \int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\ dx$

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

Let , $\ x-\frac{1}{x}=t$

$\therefore\ \ \left(1+\frac{1}{x^2}\right)\ dx=dt$

$\therefore\ \ I_1=\ \int\frac{1}{t^2+2}\ dt$

$=\ \frac{1}{\sqrt2}\tan^{-1}{\frac{t}{\sqrt2}}$

$=\ \frac{1}{\sqrt2}\tan^{-1}{\left(\frac{x-\frac{1}{x}}{\sqrt2}\right)}$

$=\frac{1}{\sqrt2}\tan^{-1}{\frac{x^2-1}{\sqrt2x}}----\left(2\right)$

Now,

$\therefore\ \ I_2=\ \int\frac{x^2-1}{x^4+1\ }dx$

$=\ \int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\$

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

Let, $\ x+\frac{1}{x}=y$

$\therefore\ \ \left(1-\frac{1}{x^2}\right)\ dx=dy\ and \ \int\frac{1}{y^2-2} dy$

$\therefore\ \ I_2=\ \frac{1}{2\sqrt2}\log{\left|\frac{y-\sqrt2}{y+\sqrt2}\right|}$

$=\ \frac{1}{2\sqrt2}\log{\left|\frac{x+\frac{1}{x}-\sqrt2}{x+\frac{1}{x}+\sqrt2}\right|}$

$=\ \frac{1}{2\sqrt2}\log{\left|\frac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}\right|}-----\left(3\right)$

from (1), (2) & (3)

$I=\ \frac{1}{2\sqrt2}\tan^{-1}{\frac{x^2-1}{\sqrt2x}}+\frac{1}{4\sqrt2}\log{\left|\frac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}\right|}+c$

Click here to see next pattern

Leave a Reply Cancel reply