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

Pattern - 43(iii) of Indefinite Integration

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

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⋆ Pattern – 43(iii) :

Integrals of differential binomials and chebyslev’s criterion

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

(iii) If $\frac{m+1}{n}$ is an integer and p is rational number , put $\left(a+bx^n\right)=t^k$ , k = denominator of p

iii) Find : $\int x^\frac{-1}{3}\ \left(1+x^\frac{1}{3}\right)^\frac{1}{4}\ dx$

Solution : Let,

I= $\ \int x^\frac{-1}{3}\ \left(1+x^\frac{1}{3}\right)^\frac{1}{4}\ dx$

Here, m = $\frac{-1}{3}$ , n = $\frac{1}{3}$ , p = $\frac{1}{4}$

Here, p is rational number.

$\therefore\ \ \frac{m+1}{n}=\ \frac{\frac{-1}{3}+1}{\frac{1}{3}}$ = $\ \frac{\ \ \frac{2}{3}\ \ }{\frac{1}{3}}=2\in I$

Here, k = 4

Let, 1 + $x^\frac{1}{3}$ = $t^4$ &

∴ $x^\frac{1}{3}=\ \left(t^4-1\right)$

∴ x = $\left(t^4-1\right)^3$

dx = 3 $\left(t^4-1\right)^2\times 4\ t^3\ dt$

$\therefore\ \ I=\ \int\frac{1}{\left(t^4-1\right)}\times \ \left(t^4\right)^\frac{1}{4}\times 12\left(t^4-1\right)^2\ t^3\ dt$

$=12\ \int t^4\ \left(t^4-1\right)\ dt$

$=12\ \int\left(t^8-t^4\right)\ dt$

$=12\ \left(\frac{t^9}{9}-\ \frac{t^5}{5}\right)+c$

$=4+2\ \frac{\left(5t^9-9t^5\right)}{45}+c$

$=\ \frac{4}{15}\ t^5\ \left(5t^4-9\right)+c$

$=\ \frac{4}{15}\left[\left(1+x^\frac{1}{3}\right)^\frac{1}{4}\right]^5\times\ \left[5\left(1+x^\frac{1}{3}\right)\ \right]+c$

$=\ \frac{4}{15}\ \left(1+x^\frac{1}{3}\right)^\frac{5}{4}\ \left(5x^\frac{1}{3}-4\right)\ +c$

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