${\int }_0^1{\tan }^{-1}\left(x\right) \mathit{\ⅆ}x$

$\={\left[x {\tan }^{-1}\left(x\right)\right]}_{x\=0}^{x\=1}-{\int }_0^1\frac{x}{1\+x^2} \mathit{\ⅆ}x$

$\=\left({\tan }^{-1}\left(1\right)-0\right)-{\int }_0^1\frac{x}{1\+x^2} \mathit{\ⅆ}x$

$\=\frac{\mathrm{\π}}{4}-{\int }_{x\=0}^{x\=1}\frac{1\/2}{y} \mathit{\ⅆ}y$

(The substitution $y\=1\+x^2$ is made so that $\mathrm{dy}\=2 x \mathrm{dx}$ and $x \mathrm{dx}\=\mathrm{dy}\/2$. The limits are left in terms of $x$.)

$\=\frac{\mathrm{\π}}{4}-{\left[\frac{\ln \left(y\right)}{2}\right]}_{x\=0}^{x\=1}$

(The antiderivative is found in terms of $y$.)

$\=\frac{\mathrm{\π}}{4}-{\left[\frac{\ln \left(1\+x^2\right)}{2}\right]}_{x\=0}^{x\=1}$

(Revert from $y$ to $1\+x^2$.)

$\=\frac{\mathrm{\π}}{4}-\left(\frac{\ln \left(2\right)}{2}-\frac{\ln \left(1\right)}{2}\right)$

$\=\frac{\mathrm{\π}}{4}-\left(\frac{\ln \left(2\right)}{2}-0\right)$

$\=\frac{\mathrm{\π}}{4}-\frac{\ln \left(2\right)}{2}$

Evaluation of the integral

${\int }_0^1{\tan }^{-1}\left(x\right) \mathit{\ⅆ}x$ = $\frac{1}{4}\mathrm{\π}-\frac{1}{2}\ln \left(2\right)$$$

Annotated stepwise solution

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${\int }_0^1{\tan }^{-1}\left(x\right) \mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & {\int }_0^1\arctan \left(x\right)\ⅆx\\ \text{▫}& & \text{1. Apply integration by Parts}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{Parts}}\text{rule}\\ & & \int u\ⅆv=uv-\left(\int v\ⅆu\right)\\ & \text{◦}& \text{First part}\\ & & u=\arctan \left(x\right)\\ & \text{◦}& \text{Second part}\\ & & \mathrm{dv}=1\\ & \text{◦}& \text{Differentiate first part}\\ & & \mathrm{du}=\frac{\ⅆ}{\ⅆx}arctan\left(x\right)\\ & & \mathrm{du}=\frac{1}{x^2+1}\\ & \text{◦}& \text{Integrate second part}\\ & & v=\int 1\ⅆx\\ & & v=x\\ & & \int \arctan \left(x\right)\ⅆx=\arctan \left(x\right)x-\left(\int \frac{x}{x^2+1}\ⅆx\right)\\ & & \text{This gives:}\\ & & \frac{\mathrm{\pi }}{4}-\left({\int }_0^1\frac{x}{x^2+1}\ⅆx\right)\\ \text{▫}& & \text{2. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=x^2+1\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\sqrt{u-1}\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{dx}=\frac{\mathrm{du}}{2\sqrt{u-1}}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & {\int }_0^1\frac{x}{x^2+1}\ⅆx={\int }_1^2\frac{1}{2u}\ⅆu\\ & & \text{This gives:}\\ & & \frac{\mathrm{\pi }}{4}-\left({\int }_1^2\frac{1}{2u}\ⅆu\right)\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{1}{2u}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{1}{2u}\ⅆu=\frac{\left(\int \frac{1}{u}\ⅆu\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\mathrm{\pi }}{4}-\frac{\left({\int }_1^2\frac{1}{u}\ⅆu\right)}{2}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{reciprocal}}\text{rule to the term}\int \frac{1}{u}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{reciprocal}}\text{rule}\\ & & \int \frac{1}{u}\ⅆu=\ln \left(u\right)\\ & \text{◦}& \text{Apply limits of definite integral}\\ & & \genfrac{}{}{0ex}{}{\ln \left(u\right)}{\phantom{u=2}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u\right)}}{u=2}-\left(\genfrac{}{}{0ex}{}{\ln \left(u\right)}{\phantom{u=1}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u\right)}}{u=1}\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\mathrm{\pi }}{4}-\frac{\ln \left(2\right)}{2}\end{array}$