Navsari 396445, Gujarat, INDIA.
+91 94281 63811

Blog

Parin Sir > Mathematical Proofs > Indefinite Integration > Pattern - 43(iv) of Indefinite Integration > Pattern - 43(iv) of Indefinite Integration

Pattern - 43(iv) of Indefinite Integration

February 24, 2021
Category: Pattern - 43(iv) of Indefinite Integration

No Comments

⋆ Pattern – 43 (iv)

Integrals of differential binomials and chebyslev’s criterion

$\int x^m\left(a+bx^n\right)^p\ dx$ : p ≠ 0

(iv) If $\frac{m+1}{n}$ +p is an integer, put $\frac{a+bx^n}{x^n}=t^k$ , & p is rational number , k = denominator of p.

(iv) find : $\int \frac{1}{\sqrt{\left(1+x^2\right)^3}}\ dx$

Solution. Let,

I= $\ \int\frac{1}{\sqrt{\left(1+x^2\right)^3}}\ dx$

$=\ \int{x^0\times\left(1+x^2\right)^\frac{-3}{2}\ dx}$

Here, m = 0, n= 2, p = $\frac{-3}{2}$

P is rational number now, $\frac{m+1}{n}=\ \frac{0+1}{2}=\frac{1}{2}$

$\therefore\ \ \frac{m+1}{n}+p$ = $\ \frac{1}{2}+\ \left(\frac{-3}{2}\right)$ = -2 is an integer.

Here, k = 2

Let, $\frac{1+x^2}{x^2}=t^2$

$\therefore\ \ \frac{1}{x^2}+1=t^2$

$\therefore\ \ x^{-2}=t^2-1$

$\therefore\ -2x^{-3}dx = 2t dt$

$\therefore\ \ \frac{-1}{x^3}dx=t\ dt$

$\therefore\ \ \frac{1}{x^3}\ dx=\ -t dt$

$\therefore\ \ I=\ \int{\left(x^2t^2\right)^\frac{-3}{2}\times\ \left(-tx^3\right)\ dt}$

$=\ -\int{x^{-3}\times t^{-3}\times t\times x^3\ dt}$

$=\ -\int{t^{-2}\ dt}$

$=\ -\frac{t^{-1}}{-1}+c$

$=\ \frac{1}{t}+c$

$=\ \frac{1}{\sqrt{\frac{1+x62\ }{x^2}}}+c$

$=\ \sqrt{\frac{x^2}{1+x^2}}+c$

Thank you…

Leave a Reply Cancel reply