Question

*20 POINT* please answer fast​

Answer 1

$\implies\displaystyle \int \dfrac{1}{1 - {x}^{6} }dx$

we can also write the above expression as :-

$\implies\displaystyle \int \dfrac{1}{1 - {( {x}^{3}) }^{2} }dx$

⟹ (x³)² = x⁶

Now we know that :-

$\implies\displaystyle \int \dfrac{1}{ {a}^{2} - {y}^{2} }dy = \frac{1}{2a} log( \dfrac{a + y}{a - y} ) + c$

We will use the above formula :

$\implies\displaystyle \int \dfrac{1}{ {1}^{2} - {( {x}^{3}) }^{2} }dx$

here y = x³ and a = 1

$\implies\displaystyle \dfrac{1}{2} log( \dfrac{1 + {x}^{3} }{1 - {x}^{3} } ) + c$

Where c = constant

Answer :

$\displaystyle \dfrac{1}{2} log( \dfrac{1 + {x}^{3} }{1 - {x}^{3} } ) + c$

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

( Also check attachment for more Formulae )

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Answer 2

Given,

The function is

• $\tt y = \frac{1}{1 - {x}^{6} }$

Integrating wrt to x , we get

$\tt \implies \int{ \frac{1}{1 - {x}^{6} } } \: dx$

$\tt \implies \int{ \frac{1}{ {(1)}^{2} - {( {x}^{3} )}^{2} } } \: dx$

$\tt\implies \frac{1}{2 \times 1} log( \frac{1 + {x}^{3} }{1 - {x}^{3} } ) + c$

$\tt \implies \frac{1}{2 } log( \frac{1 + {x}^{3} }{1 - {x}^{3} } ) + c$

Remmember :

$\star \: \: \tt \int \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} log( \frac{x - a}{x + a} ) + c$

$\star \: \: \tt \int \frac{dx}{ {a}^{2} - {x}^{2} } = \frac{1}{2a} log( \frac{a + x}{a -x} ) + c$

$\star \: \: \tt \int{ \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } } = log(x + \sqrt{ {x}^{2} + {a}^{2} } ) + c$

$\star \: \: \tt \int{ \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } } = log(x + \sqrt{ {x}^{2} - {a}^{2} } ) + c$

$\star \: \: \tt \int{ \frac{dx}{ {x}^{2} + {a}^{2} } } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c$

$\star \: \: \tt \int{ \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } } = {sin}^{ - 1} \frac{x}{a} + c$