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Problem Define S ( x ) = ∫ x arctan x 1 + t 4 d t S(x)=\displaystyle\int_{x}^{\arctan x} \sqrt{1+t^{4}}\,dt S ( x ) = ∫ x a r c t a n x 1 + t 4 d t . Which of the following is equal to S ′ ( x ) S'(x) S ′ ( x ) ?
a) 1 + ( arctan x ) 4 1 + x 2 + 1 + x 4 \dfrac{\sqrt{1+(\arctan x)^{4}}}{1+x^{2}}+\sqrt{1+x^{4}} 1 + x 2 1 + ( arctan x ) 4 + 1 + x 4
b) 1 + ( arctan x ) 4 1 + x 2 − 1 + x 4 1 + x 2 \dfrac{\sqrt{1+(\arctan x)^{4}}}{1+x^{2}}-\dfrac{\sqrt{1+x^{4}}}{1+x^{2}} 1 + x 2 1 + ( arctan x ) 4 − 1 + x 2 1 + x 4
c) 1 + ( arctan x ) 4 1 + x 2 − 1 + x 4 \dfrac{\sqrt{1+(\arctan x)^{4}}}{1+x^{2}}-\sqrt{1+x^{4}} 1 + x 2 1 + ( arctan x ) 4 − 1 + x 4
d) x 1 + ( arctan x ) 4 1 + x 2 − 1 + x 4 \dfrac{x\sqrt{1+(\arctan x)^{4}}}{1+x^{2}}-\sqrt{1+x^{4}} 1 + x 2 x 1 + ( arctan x ) 4 − 1 + x 4
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