Question

integration of x^2 -4 /x^2 +4 limit 0 to 4​

Answer 1

$\displaystyle\sf\int\limits_0^4\:\dfrac{x^2-4}{x^2+4}\:dx$

$\displaystyle\sf = \int\limits_0^4 \left( 1- \dfrac{8}{x^2+4}\right)dx$

$\displaystyle\sf = -8\:\int\limits_0^4 \dfrac{1}{x^2+4} dx + \int\limits_0^4 \:1\:\:dx$

$\displaystyle\sf = -8\:\int\limits_0^4\dfrac{1}{4\left(\dfrac{x^2}{4}+1\right)} dx + \int\limits_0^4\:1\;\;dx$

$\displaystyle\sf = -2\:\int\limits_0^4\dfrac{1}{\dfrac{x^2}{4}+1} dx + \int\limits_0^4\:1\:\;dx$

• substitute u = x/2

$\displaystyle\sf\implies du = \dfrac{1}{2}dx$

this gives us a new lower limit

• 0/2 = 0

also a new upper limit

• 4/2 = 2

$\displaystyle\sf = -4 \int\limits_{\red{0}}^{\blue{2}} \dfrac{1}{u^2+1}du + \int\limits_{\red{0}}^{\green{4}}\:1\;\;dx$

$\displaystyle\sf = \bigg[ -4\:tan^{-1}\:u\bigg]^2_0 + \int\limits_0^4\:2\:\:dx$

$\displaystyle\sf = -4\:tan^{-1}\:2+[x]^4_0$

$\boxed{\displaystyle\sf = 4 - 4\:tan^{-1} (2)}$

Answer 2

$\huge{\mathfrak{\green{Answer:-}}}$

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