Evaluate $\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x$

Evaluate the following integral: $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$

Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{x}{(1+x^4)^{1/4}}$ and $g(x)=f(f(f(f(x))))$. Then find the value of $18 \int_0^{\sqrt{2\sqrt{5}}} x^3 g(x) dx$.

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$,respectively with the positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}$,where $\alpha$ and $\beta$ are integers,then the value of $\alpha+\beta$ equals:

$\int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx$ is equal to

$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to