Stacie Worsley

2022-01-03

Determine the following indefinite integral.
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx$

Jim Hunt

Step 1
Indefinite integral is basically an integral without upper and lower bounds, i.e. its boundaries are not set
Step 2
To find: $\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx$
Now, taking ${x}^{2}$ common from the numerator and the denominator and simplifying:
$\int \frac{1}{{x}^{2}+1}dx$
Let $x=\mathrm{tan}u$
$dx={\mathrm{sec}}^{2}udu$
Substitute
$=\int \frac{1}{1+{\mathrm{tan}}^{2}u}\cdot {\mathrm{sec}}^{2}udu$
Now, as we know that $1+{\mathrm{tan}}^{2}u={\mathrm{sec}}^{2}u$
$=\int \frac{1}{{\mathrm{sec}}^{2}u}\cdot {\mathrm{sec}}^{2}udu$
$=\int 1\cdot du$
Integrating,
=u+c
Now, as $x=\mathrm{tan}u$
$u={\mathrm{tan}}^{-1}x$
Substituting value of u
$={\mathrm{tan}}^{-1}x+c$
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx={\mathrm{tan}}^{-1}x+c$

alkaholikd9

We need to evaluate the indefinite integral
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx$
To do this we can use following rules,
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C⇒\left(1\right)$
and,
where C is a real constant. Then,
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx=\int \frac{{x}^{2}}{{x}^{2}\left({x}^{2}+1\right)}dx$
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx=\int \frac{1}{{x}^{2}+1}dx$
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx={\mathrm{tan}}^{-1}x+C$
where, C is the constant of integration.
Result:
The value of the given indefinite integral is,
$\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx={\mathrm{tan}}^{-1}x+C$
where C is the constant of integration.

karton

$\phantom{\rule{0ex}{0ex}}\text{Given:}\phantom{\rule{0ex}{0ex}}\int \frac{{x}^{2}}{{x}^{4}+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}\text{Factor the expression}\phantom{\rule{0ex}{0ex}}\int \frac{{x}^{2}}{{x}^{2}×\left({x}^{2}+1\right)}dx\phantom{\rule{0ex}{0ex}}\text{Reduce the fraction}\phantom{\rule{0ex}{0ex}}\int \frac{1}{{x}^{2}+1}dx\phantom{\rule{0ex}{0ex}}\text{Evaluate}\phantom{\rule{0ex}{0ex}}\frac{1}{1}×\mathrm{arctan}\left(\frac{x}{1}\right)\phantom{\rule{0ex}{0ex}}\mathrm{arctan}\left(x\right)\phantom{\rule{0ex}{0ex}}\text{Solution:}\phantom{\rule{0ex}{0ex}}\mathrm{arctan}\left(x\right)+C$