Question

plz answer this question guys​

Answer 1

Given :

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

Solution :

$\displaystyle{\sf \implies \int \dfrac{1}{x^3+1}\:dx}$

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

Using partial differentiation,

$\displaystyle{\sf \implies \dfrac{1}{(x+1)(x^2-x+1)}\:dx= \dfrac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}}$

$\displaystyle{\sf \implies \dfrac{1}{(x+1)(x^2-x+1)}\:dx= \dfrac{Ax^2-Ax+A+Bx^2+Bx+Cx+C}{(x+1)(x^2-x+1)}}$

$\implies \displaystyle{\sf (A+B)x^2+(B+C-A)x+A+C=1}$

$\displaystyle{\implies \sf A+B=0 }\\\\{\sf \implies B=-A}\\\\{\implies \sf B+C-A=0}\\\\{\implies \sf A+C=1 }\\\\{\implies \sf C = 2A } \\\\\begin{array}{cc}\boxed{\sf C = \dfrac{2}{3}} &\boxed{\sf A = \dfrac{1}{3}}\end{array}$

$\displaystyle{\Rightarrow \sf \dfrac{1}{(x+1)(x^2-x+1)}=\dfrac{1}{3(x+1)}+\dfrac{\Big(\dfrac{-1}{3}x+\dfrac{2}{3}\Big)}{x^2-x+1}}$

$\displaystyle{\Rightarrow \sf \int \dfrac{dx}{(x+1)(x^2-x+1)}=\int \dfrac{dx}{3(x+1)}-\int \dfrac{-x+2}{3(x^2-x+1)}dx}$

$\displaystyle{\Rightarrow \sf \dfrac{1}{3}\bigg[\int\dfrac{1}{x+1}dx+\int\dfrac{-x+2}{x^2-x+1}dx\bigg]}$

$\displaystyle{\boxed{\sf x+1=t\Rightarrow dt = dx\left | {{x^2-x+1=u} \atop {(2x-1)dx=dv}} \right. }}$

$\displaystyle{\Rightarrow \sf \Bigg[\int \dfrac{1}{t}(dt)+\dfrac{1}{2}\int\dfrac{-2x+1-1}{x^2-x+1}dx+\int \dfrac{2}{x^2-x+1}dx\Bigg]}$

$\displaystyle{\Rightarrow \sf \Bigg[in\: t - \dfrac{1}{2}\int\dfrac{2x-1}{x^2-x+1}dx-\int \dfrac{1}{2(x^2-x+1)}dx+\int \dfrac{2}{x^2-x+1}dx \Bigg]}$

$\displaystyle{\Rightarrow \sf \dfrac{1}{3}\Bigg[in\:|x+1|-\dfrac{1}{2}\int\dfrac{dv}{u}+\int \dfrac{-1+4}{2(x^2-x+1)}dx\Bigg]}$

$\displaystyle{\Rightarrow \sf \dfrac{1}{3}\Bigg[in\:|x+1|-\dfrac{1}{2}\:in\:(u)+\dfrac{3}{2}\int \dfrac{1}{x^2-x+1}dx\Bigg]}$

$\displaystyle{\Rightarrow \sf \dfrac{1}{3}\Bigg[in\:|x+1|-\dfrac{1}{2}\:in\:|x^2-x+1|+\dfrac{3}{2}\int\dfrac{1}{x^2-2\times x \times \dfrac{1}{2}+\Big(\dfrac{1}{2}\Big)^2+1-\Big(\dfrac{1}{2}\Big)^2}\:dx\Bigg]}$

$\displaystyle{\sf \Rightarrow \dfrac{1}{3}\Bigg[in\:|x+1|-\dfrac{1}{2}\:in\:|x^2-x+1|+\dfrac{3}{2} \int \dfrac{dx}{\Big(x-\dfrac{1}{2}\Big)^2+\Big(\dfrac{\sqrt{3}}{2}\Big)^2}\Bigg]}$

$\displaystyle{\sf \Rightarrow \dfrac{1}{3}\Bigg[in\:|x+1|-\dfrac{1}{2}\:in\:|x^2-x+1|+\dfrac{3}{2}.\dfrac{2}{\sqrt{3}}.tan^{-1}.\dfrac{x-1}{\dfrac{\sqrt{3}}{2}}\Bigg]}$

Required Answer :

$\displaystyle{\boxed{\boxed{\sf \int \dfrac{dx}{x^3+1}=\dfrac{1}{3}\Bigg[in \:|x+1|-\dfrac{1}{2}\:in\:|x^2-x+1|+\sqrt{3}\:tan^{-1}\Big(\dfrac{2(x-1)}{\sqrt{3}}\Big)+C\Bigg]}}}$