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Question Number 123314 by aurpeyz last updated on 24/Nov/20

$$\int \frac{8}{2x^2+3x+5}dx$$

Answered by mathmax by abdo last updated on 24/Nov/20

$$\mathrm{I}=8\int \frac{\mathrm{dx}}{2({\mathrm{x}}^2+\frac{3}{2}\mathrm{x}+\frac{5}{2})}=4\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+2.\frac{3}{4}\mathrm{x}+\frac{9}{16}+\frac{5}{2}-\frac{9}{16}}$$$$=4\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{3}{4})}^2+\frac{31}{16}}{=}_{\mathrm{x}+\frac{3}{4}=\frac{\sqrt{31}}{4}\mathrm{u}}4\times \frac{16}{31}\int \frac{1}{{\mathrm{u}}^2+1}.\frac{\sqrt{31}}{4}\mathrm{du}$$$$=\frac{16}{\sqrt{31}}\arctan \left(\mathrm{u}\right)+\mathrm{C}=\frac{16}{\sqrt{31}}\arctan \left(\frac{4\mathrm{x}+3}{\sqrt{31}}\right)+\mathrm{C}$$

Commented by aurpeyz last updated on 25/Nov/20

$$thankyou$$

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