Question
Evaluate the following integrals:$\int_{0}^\limits{1}\tan^{-1}\text{x dx}$

Answer

Let $\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$ Then,$\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
Integrating by parts,$\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\frac{1}{2}\big[\log\big(\text{x}^2+1\big)\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0-\frac{1}{2}\log2+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-\frac{1}{2}\log2$

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