Evaluate the following integrals: x^3 (x^2+1)^3 dx - Vidyadip

Question

Evaluate the following integrals: $\int\frac{\text{x}^3}{(\text{x}^2+1)^3}\text{dx}$

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Answer

Let I $=\int\frac{\text{x}^3}{(\text{x}^2+1)^3}\text{dx}\ ....(1)$
Let $1 + x^2 = t $ then,
$d(1 + x^2) = dt$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Puttting $1 + x^2 = t$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation $(1),$ we get
$\text{I}=\int\frac{\text{x}^2}{\text{t}^3}\times\frac{\text{dt}}{2}$
$=\frac{1}{2}\int\frac{(\text{t}-1)}{\text{t}^3}\text{dt}\ \ [\because1+\text{x}^2=\text{t}]$
$=\frac{1}{2}\int\Big[\Big(\frac{\text{t}}{\text{t}^3}-\frac{1}{\text{t}^3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\int\big(\text{t}^{-2}-\text{t}^{-3}\big)\text{dt}$
$=\frac{1}{2}\Big[-1\text{t}^{-1}-\frac{\text{t}^{-2}}{-2}\Big]+\text{C}$
$=\frac{1}{2}\Big[-\frac{1}{\text{t}}+\frac{1}{2\text{t}^2}\Big]+\text{C}$
$=-\frac{1}{2\text{t}}+\frac{1}{4\text{t}^2}+\text{C}$
$=-\frac{1}{2(1+\text{x}^2)}+\frac{1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2(1+\text{x}^2)+1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2-2\text{x}^2+1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2\text{x}^2-1}{4(1+\text{x}^2)^2}+\text{C}$
$=-\frac{(1+2\text{x}^2)}{4(\text{x}^2+1)^2}+\text{C}$
$\therefore\text{I}=-\frac{(1+2\text{x}^2)}{4(\text{x}^2+1)^2}+\text{C}$

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