∫dxx2x4+13/4=Ax4+1x4B+C

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

1. A

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

2. B

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

3. C

$\mathrm{A}=\frac{1}{2},\mathrm{B}=1$

4. D

$\mathrm{A}=-\frac{1}{2},\mathrm{B}=-1$

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### Solution:

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

$\begin{array}{r}\mathrm{Put}1+{\mathrm{x}}^{-4}=\mathrm{t}⇒\frac{-4}{{\mathrm{x}}^{5}}\mathrm{dx}=\mathrm{dt}⇒-\frac{1}{4}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{3/4}}\\ =\frac{1}{4}\int {\mathrm{t}}^{-3/4}\mathrm{dt}=-\frac{1}{4}\cdot \frac{{\mathrm{t}}^{1/4}}{1/4}+\mathrm{C}\\ =-{\left(1+\frac{1}{{\mathrm{x}}^{4}}\right)}^{1/4}+\mathrm{C}=-{\left(\frac{1+{\mathrm{x}}^{4}}{{\mathrm{x}}^{4}}\right)}^{1/4}+\mathrm{C}\end{array}$

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