Question

Integrate the given function :-

$\int \dfrac{ {x}^{4} }{1 + {x}^{2} } \: dx$
​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to integrate the given function.

$\displaystyle = \rm \int \dfrac{ {x}^{4} }{1 + {x}^{2} } \: dx$

We can write it as:

$\displaystyle = \rm \int \dfrac{ {x}^{4} - 1 + 1 }{1 + {x}^{2} } \: dx$

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

$\displaystyle = \rm \int \bigg( \dfrac{1}{1 + {x}^{2} } + {x}^{2} - 1 \bigg) \: dx$

$\displaystyle = \rm \int \dfrac{1}{1 + {x}^{2} } \: dx + \int {x}^{2} \: dx - \int 1 \: dx$

We know that:

$\displaystyle \rm: \longmapsto \int \dfrac{1}{1 + {x}^{2} } \: dx = \arctan(x) + C$

$\displaystyle \rm: \longmapsto \int {x}^{n}\: dx = \dfrac{ {x}^{n + 1} }{n + 1} + C$

Therefore, we get:

$\displaystyle = \rm \arctan(x) + \dfrac{ { x }^{2 + 1} }{2 + 1} - x + C$

$\displaystyle = \rm \dfrac{ { x }^{3} }{3} - x + \arctan(x) + C$

Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf kx+C\\ \\ \sf sin(x)&\sf-cos(x)+C\\ \\ \sf cos(x)&\sf sin(x)+C\\ \\ \sf{sec}^{2}(x)&\sf tan(x)+C\\ \\ \sf{cosec}^{2}(x)&\sf-cot(x)+C\\ \\ \sf sec(x)\ tan(x)&\sf sec(x)+C\\ \\ \sf cosec(x)\ cot(x)&\sf-cosec(x)+C\\ \\ \sf tan(x)&\sf log(sec(x))+C\\ \\ \sf\dfrac{1}{x}&\sf log(x)+C\\ \\ \sf{e}^{x}&\sf{e}^{x}+C\\ \\ \sf x^{n},n\neq-1&\sf\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

ANSWER:

To Integrate:

• (x⁴)/(1+x²)

Solution:

We are given that,

$\displaystyle\implies\int\dfrac{x^4}{1+x^2}\:dx$

$\displaystyle\implies\int f(x)\:dx$

We will first simplify the function. So,

$\implies f(x)=\dfrac{x^4}{1+x^2}$

Adding and subtracting 1,

$\implies f(x)=\dfrac{x^4-1+1}{1+x^2}$

$\implies f(x)=\dfrac{(x^4-1)+1}{x^2+1}$

$\implies f(x)=\dfrac{(x^2-1)(x^2+1)+1}{x^2+1}$

Separating the terms,

$\implies f(x)=\dfrac{(x^2-1)(x^2+1)}{x^2+1}+\dfrac{1}{x^2+1}$

Simplifying,

$\implies f(x)=x^2-1+\dfrac{1}{x^2+1}$

So,

$\displaystyle\implies\int f(x)\:dx$

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

By separating the terms,

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

We know that,

$\displaystyle\hookrightarrow \int x^n\:dx=\dfrac{x^{n+1}}{n+1}$

$\displaystyle\hookrightarrow \int 1\:dx=x$

And,

$\displaystyle\hookrightarrow \int \dfrac{1}{x^2+1}\:dx=\arctan(x)$

So,

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

$\implies\dfrac{x^{2+1}}{2+1}-x+\arctan(x)+C$

$\implies\dfrac{x^3}{3}-x+\arctan(x)+C$

Hence,

$\displaystyle\bf\implies\int\dfrac{x^4}{1+x^2}\:dx=\dfrac{x^3}{3}-x+arctan(x)+C$

OR

$\displaystyle\bf\implies\int\dfrac{x^4}{1+x^2}\:dx=\dfrac{x^3}{3}-x+tan^{-1}(x)+C$