Antiderivatives and definite integrals

Question

Compute the following integral: $$ I=\int_{0}^{1}\bigg(\int_{1}^{x}\frac{1}{1+t^2}dt\bigg)dx $$ Motivation: From time to time, i intend to post here challenging questions -together with proposed answers/solutions- which i frequently give for practise to some of my students (in an introductory calculus course).
I am doing this with the understanding that this is not only accepted but furthermore encouraged by the community. The benefits can be twofold: on the one hand this contributes to building a library with well-posed questions and reliable answers (to be used in-class or online) and on the other hand i always hope that this may result in improving (or sometimes correcting) the proposed questions/answers and discovering new approaches/solutions from other users as well.

Answer 1

Let $f(t)=\frac{1}{1+t^2}$, $t\in\mathbb{R}$ and $F(x)=\int_{1}^{x}\frac{1}{1+t^2}dt=\int_{1}^{x}f(t)dt$, $t\in\mathbb{R}$.

Consequently, $F'(x)=f(x)$, $x\in\mathbb{R}$ and: $$ I=\int_{0}^{1}\bigg(\int_{1}^{x}\frac{1}{1+t^2}dt\bigg)dx=\int_{0}^{1}F(x)dx=\int_{0}^{1}(x)'F(x)dx= $$ $$ =[xF(x)]_{0}^{1}-\int_{0}^{1}xF'(x)dx=F(1)-\int_{0}^{1}xf(x)dx=0-\int_{0}^{1}\frac{x}{1+x^2}dx= $$ $$ =-\frac{1}{2}\int_{0}^{1}\frac{(1+x^2)'}{1+x^2}dx=-\frac{1}{2}[\ln(1+x^2)]_{0}^{1}=-\frac{\ln 2}{2} $$

Answer 2

$\begin{align}\displaystyle \int_0^1\int_1^x\frac{dt}{1+t^2}\cdot dx&=\int_0^1\left(\arctan(x)-\frac{\pi}{4}\right)dx\\&=\bigg[x\arctan(x)-\frac 12\ln(1+x^2)-\frac{\pi}4x\bigg]_0^1=-\frac12\ln(2)\end{align}$