∫dxx2x4+13/4=Ax4+1x4B+C - Infinity Learn by Sri Chaitanya

A
$A = - 1, B = \frac{1}{4}$
B
$A = 1, B = - \frac{1}{4}$
C
$A = \frac{1}{2}, B = 1$
D
$A = - \frac{1}{2}, B = - 1$

Solution:

$I = \int \frac{dx}{x^{2} {(x^{4} + 1)}^{3 / 4}} = \int \frac{dx}{x^{2} \cdot x^{3} {(1 + \frac{1}{x^{4}})}^{3 / 4}}$

$\begin{aligned} Put 1 + x^{- 4} = t \Rightarrow \frac{- 4}{x^{5}} dx = dt \Rightarrow - \frac{1}{4} \int \frac{dt}{t^{3 / 4}} \\ = \frac{1}{4} \int t^{- 3 / 4} dt = - \frac{1}{4} \cdot \frac{t^{1 / 4}}{1 / 4} + C \\ = - {(1 + \frac{1}{x^{4}})}^{1 / 4} + C = - {(\frac{1 + x^{4}}{x^{4}})}^{1 / 4} + C \end{aligned}$

$Hence, A = - 1, B = \frac{1}{4}$

$\int \frac{dx}{x^{2} {(x^{4} + 1)}^{3 / 4}} = A {(\frac{x^{4} + 1}{x^{4}})}^{B} + C$

Solution:

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