<s>[INST] How can I handle the integation which contains $~ \left( 1+ t ^{2} \right) ^{2} ~$ and $~ \left( 1- t ^{2} \right) ^{2} ~~$? $$  \left( a,d  \in \mathbb R_{> 0}  \right) ~~\wedge~~ \left( a < d \right) $$
I have to find out the value of the below integral .
$$   \alpha :=  \int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2}  \right) ^{2}   }{   \left( 1+ t ^{2}  \right) \left\{ d ^{2} \left( 1+ t ^{2}  \right) ^{2} - a ^{2} \left( 1- t ^{2}  \right) ^{2}  \right\}  }  \,dt    $$
$$ = 2a \int_{0 }^{1 } \frac{ \left( 1- t ^{2}  \right) ^{2}   }{   \left( 1+ t ^{2}  \right) \left\{ d ^{2} \left( 1+ t ^{2}  \right) ^{2} - a ^{2} \left( 1- t ^{2}  \right) ^{2}  \right\}  }  \,dt  $$
My tries so far are as below .
$$  t= \tan^{}\left( x \right)  $$
$$  t : 0 ~\rightarrow~1 $$
$$  x : 0 ~\rightarrow~ \frac{\pi}{4}  $$
$$  \frac{  dt  }{ dx   }  = \sec^{2}\left( x \right)  $$
$$  1+ t ^{2} = \sec^{2}\left(  x\right)  $$
$$  \therefore ~~ \frac{  dt  }{  dx  } = \left( 1 + t ^{2}  \right)  $$
$$  \frac{  dt  }{  \left( 1+  t  ^{2}  \right)   }= dx  $$
$$  1+ t ^{2} = \sec^{2}\left( x \right) ~~ \leftarrow~~ \text{I will subtract each right and the left term  by }~~ -2 t ^{2}   $$
$$  1+ t ^{2}  -2 ^{2} = \sec^{2}\left( x \right) - 2 t  ^{2}  $$
$$  1- t ^{2} = \sec^{2}\left( x \right) - 2 \left( \sec^{2}\left( x \right) -1 \right)  $$
$$ = \sec^{2}\left( x \right) -2 \sec^{2}\left( x \right) + 2 $$
$$ = - \sec^{2}\left( x \right) +2 $$
$$ = 2- \sec^{2}\left( x \right)  $$
$$  \therefore ~~  \alpha  = 2a \int_{0 }^{\frac{\pi}{4}  } \frac{  \left( 2- \sec^{2}\left( x \right)  \right) ^{2}   }{  \left\{ d ^{2} \left( \sec^{2}\left( x \right)  \right) ^{2} -a ^{2} \left( 2 - \sec^{2}\left( x_{}  \right)   \right)^{2}  \right\}   }  \,dt   $$
I've been got stucked from here .
 [/INST] Substitute $t=\tan \frac x2$
\begin{align}
&\int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2}  \right) ^{2}   }{   \left( 1+ t ^{2}  \right) \left\{ d ^{2} \left( 1+ t ^{2}  \right) ^{2} - a ^{2} \left( 1- t ^{2}  \right) ^{2}  \right\}  }  \,dt    \\
=& \int_0^{\frac\pi2}\frac{a \cos^2x}{d^2 - a^2\cos^2x}dx
= \frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{d^2}{d^2 - a^2\cos^2x}dx\right)\\
 = &\frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{\>d(\tan x)}{\frac{d^2 - a^2 }{d^2}+ \tan^2x}\right)
=\frac\pi{2a}\left( \frac d{\sqrt{d^2-a^2}}-1\right)
\end{align}
</s>