If -x4-x-4-2x3dx-ln -x-axb-C- then absolute value of ba is where a-b are fixed contants and C is constant of integration

√((x^4+x^ 4+2))=√((x^2+1x^2)^2) ∴∫√(x^4+x^ 4+2)x^3dx=∫(x^2+1x^2x^3)dx ⇒∫(1x+1x^5)dx =ln |x|+x^ 4 4+C =ln |x| 14x^4+C ⇒ a= 14,b=4 ⇒ absolute value of ba=16 ...

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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b ± c)²

If ∫√x4+x-4+2...

Question

If $\int \frac{\sqrt{x^{4} + x^{- 4} + 2}}{x^{3}} d x = ln | x | + \frac{a}{x^{b}} + C$ , then absolute value of $(\frac{b}{a})$ is (where $a, b$ are fixed contants and $C$ is constant of integration)

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Solution

$\sqrt{(x^{4} + x^{- 4} + 2)} = \sqrt{{(x^{2} + \frac{1}{x^{2}})}^{2}}$
$∴ \int \frac{\sqrt{x^{4} + x^{- 4} + 2}}{x^{3}} d x = \int ⎛ ⎜ ⎜ ⎜ ⎝ \frac{x^{2} + \frac{1}{x^{2}}}{x^{3}} ⎞ ⎟ ⎟ ⎟ ⎠ d x$
$\Rightarrow \int (\frac{1}{x} + \frac{1}{x^{5}}) d x$
$= ln | x | + \frac{x^{- 4}}{- 4} + C$
$= ln | x | - \frac{1}{4 x^{4}} + C$
$\Rightarrow a = - \frac{1}{4}, b = 4$
$\Rightarrow$ absolute value of $\frac{b}{a} = 16$

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