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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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