Determine the following indefinite integral. ∫x2x4+x2dx

Question

Determine the following indefinite integral.  ∫x2x4+x2dx

Jim Hunt · Accepted Answer

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}} d x

Now, taking

x^{2}

common from the numerator and the denominator and simplifying:

\int \frac{1}{x^{2} + 1} d x

Let

x = \tan u

d x = \sec^{2} u d u

Substitute

= \int \frac{1}{1 + \tan^{2} u} \cdot \sec^{2} u d u

Now, as we know that

1 + \tan^{2} u = \sec^{2} u

= \int \frac{1}{\sec^{2} u} \cdot \sec^{2} u d u

= \int 1 \cdot d u

Integrating,
=u+c
Now, as

x = \tan u

u = \tan^{- 1} x

Substituting value of u

= \tan^{- 1} x + c

\int \frac{x^{2}}{x^{4} + x^{2}} d x = \tan^{- 1} x + c

alkaholikd9 · Accepted Answer

We need to evaluate the indefinite integral

\int \frac{x^{2}}{x^{4} + x^{2}} d x

To do this we can use following rules,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C \Rightarrow (1)

and,
where C is a real constant. Then,

\int \frac{x^{2}}{x^{4} + x^{2}} d x = \int \frac{x^{2}}{x^{2} (x^{2} + 1)} d x

\int \frac{x^{2}}{x^{4} + x^{2}} d x = \int \frac{1}{x^{2} + 1} d x

\int \frac{x^{2}}{x^{4} + x^{2}} d x = \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}} d x = \tan^{- 1} x + C

where C is the constant of integration.

karton · Accepted Answer

Given:∫x2x4+x2dxFactor the expression∫x2x2×(x2+1)dxReduce the fraction∫1x2+1dxEvaluate11×arctan⁡(x1)arctan⁡(x)Solution:arctan⁡(x)+C

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