Solutions - 0010-05

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u={\tan }^{-1}x& v^{\prime }=1\\ u^{\prime }={\displaystyle \frac{1}{1+x^2}}& v=x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int {\tan }^{-1}x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int {\displaystyle \frac{x}{1+x^2}}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cccc}& uv-\int u^{\prime }\cdot v& \gg \gg x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx& \text{(Exp. A)}\end{array}$$

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(3) Perform $u$-sub with $u=1+x^2$ and $du=2xdx$:

$$\begin{array}{cc}\int \frac{x}{1+x^2}dx& \gg \gg \frac{1}{2}\int \frac{2xdx}{1+x^2}\\ & \\ & \gg \gg \frac{1}{2}\int \frac{du}{u}\\ & \\ & \gg \gg \frac{1}{2}\ln |u|+C\\ & \\ & \gg \gg \frac{1}{2}\ln |1+x^2|+C\end{array}$$

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(4) Insert result in Exp. (A):

$$\begin{array}{ccc}& (A)\gg \gg x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)+C& \text{(Note B)}\end{array}$$

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Note B: We can change $|1+x^2|$ to $(1+x^2)$ because the inner expression is never negative.