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I have to find the values of $x \in \mathbb R$ so that

$$\int _0^1\frac{dt}{1+\left\{x+t\right\}}\:=\:\ln2$$

To be noted that the answer is $\mathbb R$.

$$\int _0^1\:\frac{dt}{1+\left\{x+t\right\}}\:=\int _0^1\:\frac{dt}{1+x+t-\left[x+t\right]}=\int _0^1\:\frac{dt}{1+x+t-x}=\ln2$$

I solved it like this, but I don't really know, couldn't $[x+t]$ be also $x+1$ ?

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    Observe that $\{x\}\in (-1,1)$ for all $x\in\Bbb R$. Then you can divide the integrals in parts where $\{x\}=t$ for $t\in(-1,1)$. – 2017-02-25

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By a substitution $x+t=u$, $dt = du$, we have $$ \int_0^1 \frac1{1+\{x+t\}}dt = \int_x^{x+1} \frac1{1+\{u\}}du. $$ This integral is in fact independent of $x$, so we have $$ \int_x^{x+1} \frac1{1+\{u\}}du = \int_0^1 \frac1{1+\{u\}}du = \int_0^1 \frac 1{1+u}du = \ln (1+u) \bigg\vert_0^1 = \ln 2. $$

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    Why is it independent of x ? – 2017-02-25
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    Try splitting the integral into $\int_x^{\lfloor x \rfloor +1} $ and $\int_{\lfloor x \rfloor +1}^{x+1}$. – 2017-02-25
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    @i707107 Nice change of variables and translation ! If you've time, call me to upvote you tmr since my daily vote limit reached. – 2017-02-25