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Question Number 152513 by ZiYangLee last updated on 29/Aug/21

$$\int \frac{1}{2x^2+3x+8}dx=?$$

Commented by alisiao last updated on 29/Aug/21

$$$$$$=\frac{1}{2}\int \frac{dx}{{(x+\frac{3}{4})}^2+\frac{55}{16}}=\frac{1}{2}\int \frac{dt}{t^2+\frac{55}{16}}$$$$$$$$=\frac{8}{55}\int \frac{dt}{1+{\left(\frac{4}{\sqrt{55}}t\right)}^2}=\frac{32}{55\sqrt{55}}{tan}^{-1}\left(\frac{4}{\sqrt{55}}t\right)+C$$$$$$$$=\frac{32}{55\sqrt{55}}{tan}^{-1}\left(\frac{4}{\sqrt{55}}(x+\frac{3}{4})\right)+C$$$$$$

Answered by puissant last updated on 30/Aug/21

$$=\frac{1}{2}\int \frac{1}{x^2+\frac{3}{2}+4}dx$$$$=\frac{1}{2}\int \frac{1}{{(x+\frac{3}{4})}^2-\frac{9}{16}+\frac{64}{16}}dx$$$$=\frac{1}{2}\int \frac{1}{{(x+\frac{3}{4})}^2+\frac{55}{16}}dx=\frac{1}{2}\int \frac{1}{\frac{55}{16}[{\left(\frac{4}{\sqrt{55}}(x+\frac{3}{4})\right)}^2+1]}dx$$$$=\frac{8}{55}\int \frac{1}{{\left(\frac{4}{\sqrt{55}}(x+\frac{3}{4})\right)}^2+1}dx$$$$u=\frac{4}{\sqrt{55}}(x+\frac{3}{4})\to dx=\frac{\sqrt{55}}{4}du$$$$\Rightarrow I=\frac{8}{55}\times \frac{\sqrt{55}}{4}\int \frac{du}{u^2+1}$$$$\Rightarrow I=\frac{2}{\sqrt{55}}arctan\left(u\right)+C$$$$$$$$\therefore \because Q=\frac{2}{\sqrt{55}}arctan\left(\frac{4}{\sqrt{55}}(x+\frac{3}{4})\right)+C..$$

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