Prove integral limit exists
$$\lim\limits_{t\to 0^+}\left(\int\limits_0^1\frac{dx}{(x^4+t^4)^{1/4}}+\ln t\right)$$ I try to change variable $u=1/x$ then $\displaystyle\int\limits_0^1\frac{dx}{(x^4+t^4)^{1/4}}=\frac{1}{4}\int\limits_1^\infty\frac{4u^3du}{u^4(1+u^4x^4)^{1/4}}$.
But I have no idea to continue.
The integral can be solved using the substitution $u=\dfrac{\sqrt[4]{x^4+t^4}}{x}$ and some algebra maneuvering to obtain the general result $$\lim_{t\to0^+}\frac{1}{4}\left[\ln\left(\left\vert\sqrt[4]{x^4+t^4}+x\right\vert\right)-\ln\left(\left\vert\sqrt[4]{x^4+t^4}-x\right\vert\right)-2\arctan\left(\frac{\sqrt[4]{x^4+t^4}}{x}\right)\right]\bigg\vert_{0}^{1}+\ln(t)$$ $$=\lim_{t\to0^+}\frac{1}{4}\left[\ln(\sqrt[4]{t^4+1}+1)-\ln\vert\sqrt[4]{t^4+1}-1\vert-2\arctan(\sqrt[4]{t^4+1})-\ln(t)+\ln(t)+2\cdot\frac{\pi}{2}\right]+\ln(t)$$ $$=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert\dfrac{\sqrt[4]{t^4+1}+1}{\sqrt[4]{t^4+1}-1}\right\vert+\frac{\pi}{2}\right]+\ln(t)=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert t^4\cdot\dfrac{\sqrt[4]{t^4+1}+1}{\sqrt[4]{t^4+1}-1}\right\vert\right]+\frac{\pi}{8}=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert \dfrac{\sqrt[4]{t^{20}+t^{16}}+t^4}{\sqrt[4]{t^4+1}-1}\right\vert\right]+\frac{\pi}{8}\to\dfrac{0}{0}\xrightarrow{\text{L'Hopital}}=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert \dfrac{\frac{1}{4}\left(t^{20}+t^{16}\right)^{-\frac{3}{4}}\cdot \left(20t^{19}+16t^{15}\right)+4t^3}{\frac{1}{4}\left(t^4+1\right)^{-\frac{3}{4}}\cdot 4t^3}\right\vert\right]+\frac{\pi}{8}=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert \dfrac{\frac{1}{4}\left(t^{20}+t^{16}\right)^{-\frac{3}{4}}\cdot \left(20t^{16}+16t^{12}\right)+4}{\left(t^4+1\right)^{-\frac{3}{4}}}\right\vert\right]+\frac{\pi}{8}=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert \dfrac{\frac{1}{4}\left(20t^{16}+16t^{12}\right)+4}{t^{12}}\right\vert\right]+\frac{\pi}{8}=\lim_{t\to0^+}\frac{1}{4}\left[\ln\left\vert\frac{1}{4} \left(20t^{4}+16\right)+4\right\vert\right]+\frac{\pi}{8}=\frac{1}{4}\ln(8)+\frac{\pi}{8}=\boxed{\frac{3}{4}\ln(2)+\frac{\pi}{8}}\approx0.912559467$$
Note that you needed to take the limit $\lim\limits_{x\to 0}\arctan\left(\dfrac{\sqrt[4]{x^4+t^4} }{x}\right)=\frac{\pi}{2}$ earlier on in the problem.
Although my other answer was accepted by this point, I'd like to expand about the hint given in the above comments by @Did. This only shows that it converges, not what it converges to. Once you get to the step shown in those comments, use comparisons: $$0<\lim_{t\to 0}I(t)=\int_{1}^{\infty}\dfrac{1+x-(x^4+1)^{1/4}}{x(x^4+1)^{1/4}}dx<\int_{1}^{\infty}\dfrac{1+x-(x^4)^{1/4}}{x(x^4+1)^{1/4}}dx=\int_{1}^{\infty}\dfrac{1}{x(x^4+1)^{1/4}}dx<\int_{1}^{\infty}\dfrac{1}{x(x^4)^{1/4}}dx=\int_{1}^{\infty}\dfrac{1}{x^2}dx=1$$ which converges (and it should be easy for you to show/see/calculate that).