Mathematics

$$\int { \dfrac { dx }{ 1+{ x }^{ 3 } }  }$$ 


SOLUTION
Let $$ I = \int \frac{1}{1+x^{3}}dx$$
$$ =\int  \frac{1}{(1+x)(x^{2}-x+1)}dx$$ ___ (1)
Now lets use the partial decomposition Technique
assume $$ \frac{1}{(1+x)(x^{2}-x+1)} = \frac{A}{1+x}+\frac{Bx+c}{x^{2}-x+1}$$ ___ (2)
$$ A(x^{2}-x+1)+(Bx+C)(x+1) = 1$$ ___ (3)
A+x = -1 the eq'n becomes
$$ A = \frac{1}{3}$$ __(4)
Now replace A into (3) we have
$$ x^{2}-x+1+3(Bx+C)(x+1) = 3$$
Now equating the coefficients of $$x^{2}$$ and x, and
constants, we get, $$ B = \frac{-1}{3}$$ __ (5)
$$ C = \frac{2}{3}$$ __ (6)
$$ I = \int \frac{1}{3(1+x)} dx+\int \frac{-x+2}{3(x^{2}-x+1)}dx$$
$$ = \frac{1}{3}ln(1+x)-\int \frac{2x-1}{6(x^{2}-x+1)}dx\int \frac{1}{2(x^{2}-x+1)}dx$$
$$ = \frac{1}{3}ln(1+x)-\int \frac{d(x^{2}-x+1)}{6(x^{2}-x+1)}+\int \frac{1}{2(x-\frac{1}{2})^{2}+\frac{3}{4}}dx$$
$$ = \frac{1}{3}ln (1+x) -\frac{1}{6}ln(x^{2}-x+1)+I_{1}$$ __ (7)
where $$ I_{1} = \int \frac{1}{2(x-\frac{1}{2})^{2}+\frac{3}{4}}dx$$ __ (8)
Let $$ x-\frac{1}{2} = \frac{\sqrt{3}}{2} tan(y)$$ __ (9), $$ y = arc tan (\frac{2x-1}{\sqrt{3}})$$ __ (10)
$$ dx = \frac{\sqrt{3}}{2}sec^{2}(y)dy $$ ___ (11) substitute the values of (9) & (11) into 8,
we have
$$ I_{1} = \int \frac{2\sqrt{3}sec^{2}(y)}{6sec^{2}(y)}dy = \int \frac{1}{\sqrt{3}}dy = \frac{1}{\sqrt{3}}y $$ __ (12)
Replacing y with value obtained in (10)
$$ I_{1} = \frac{1}{\sqrt{3}} arc \,tan (\frac{2x-1}{\sqrt{3}})$$ __ (13)
So, $$ I = \frac{1}{6} ln \frac{(1+x)^{2}}{x^{2}-x+1}+\frac{1}{\sqrt{3}}arc\, tan (\frac{2x-1}{\sqrt{3}})$$
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Subjective Medium Published on 17th 09, 2020
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