Question

integration of x²/4+x²​

Answer 1

Answer:

Integration of x^2/x^2 + 4

= Integral x^2+4-4/x^2 + 4

= Integral x^2+4/x^2+4 - 4(1/x^2+4)

= Integral of 1 - 4 ( 1/x^2 + 4) if integrating with respect to x

then,

Final answer is

x - 2 tan^-1 ( x/2) + c

Hope this helps you !

Answer 2

Topic:

Integration

Solution:

We need to evaluate the following integral.

$\displaystyle \int\dfrac{x^2}{4 + x^2}\, dx$

Perform some algebraic manipulation here.

$\displaystyle\longrightarrow \int\dfrac{ 4 + x^2 - 4}{4 + x^2}\, dx$

$\displaystyle\longrightarrow \int\dfrac{(4 + x^2) - (4)}{4 + x^2}\, dx$

Now we can decomposite it into two fractions.

$\displaystyle\longrightarrow \int\dfrac{4 + x^2}{4 + x^2}- \dfrac{4}{4 + x^2}\, dx$

$\displaystyle\longrightarrow \int 1 - \dfrac{4}{4 + x^2}\, dx$

$\displaystyle\longrightarrow \int \, dx -\int \dfrac{4}{4 + x^2}\, dx$

$\displaystyle\longrightarrow \int \, dx -4\int \dfrac{1}{2^2 + x^2}\, dx$

Using the following identities here.

$\boxed{\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + C}$

$\boxed{\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + C}$

Using the above formulas, we get:
$\displaystyle\longrightarrow x+ C_1 -4\left[\dfrac{1}{2}\tan^{-1}\dfrac x2\right]\, + C_2$

$\displaystyle\longrightarrow x-2\tan^{-1}\left(\dfrac x2\right) +(C_1 + C_2)$

$\displaystyle\longrightarrow x-2\tan^{-1}\left(\dfrac x2\right) +C$

This is the required answer.