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Question Number 206542 by luciferit last updated on 18/Apr/24

Answered by lepuissantcedricjunior last updated on 18/Apr/24

$$\int (3\mathit{x}+5)\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)\mathit{d}\mathit{x}=\mathit{k}$$$$\{\begin{array}{l}\mathit{u}=\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)\\ {\mathit{v}}^{\prime }=(3\mathit{x}+5)\end{array}=>\{\begin{array}{l}{\mathit{u}}^{\prime }=\frac{1}{1+{\mathit{x}}^2}\\ \mathit{v}=\frac{3}{2}{\mathit{x}}^2+5\mathit{x}\end{array}$$$$\mathit{k}=(\frac{3}{2}{\mathit{x}}^2+5\mathit{x})\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)-\frac{3}{2}\int \frac{{\mathit{x}}^2+1+\frac{10}{3}\mathit{x}-1}{1+{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$\mathit{k}=(\frac{3}{2}{\mathit{x}}^2+5\mathit{x})\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)-\frac{3}{2}\mathit{x}-\frac{5}{2}\mathit{l}\mathit{n}(1+{\mathit{x}}^2)+\frac{3}{2}\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)+\mathit{c}$$$$\mathit{k}=\frac{3}{2}({\mathit{x}}^2+\frac{10}{3}\mathit{x}+1)\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)-\frac{3}{2}\mathit{x}-\frac{5}{2}\mathit{l}\mathit{n}(1+{\mathit{x}}^2)+\mathit{c}$$

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