Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 35117 by math1967 last updated on 15/May/18

$$\int \frac{2x+3}{x^4-3x-2}dx$$

Answered by MJS last updated on 16/May/18

$$\int \frac{2x+3}{x^4-3x-2}dx=\int \frac{2x+3}{(x^2-x-1)(x^2+x+2)}dx=$$$$=\int \frac{dx}{x^2-x-1}-\int \frac{dx}{x^2+x+2}=$$$$$$$$\int \frac{dx}{x^2-x-1}=\int \frac{4}{4x^2-4x-4}dx=$$$$=\int \frac{4}{(2x-\sqrt{5}-1)(2x-\sqrt{5}+1)}dx=$$$$=\frac{2\sqrt{5}}{5}(\int \frac{dx}{(2x-\sqrt{5}-1)}-\int \frac{dx}{(2x-\sqrt{5}+1)})=$$$$=\frac{\sqrt{5}}{5}\ln \left(\frac{\mid 2x-\sqrt{5}-1\mid }{\mid 2x-\sqrt{5}+1\mid }\right)$$$$$$$$\int \frac{dx}{x^2+x+2}=\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{7}{4}}=$$$$[u=\frac{\sqrt{7}}{7}(2x+1)\to dx=\frac{\sqrt{7}}{2}u]$$$$=\frac{2\sqrt{7}}{7}\int \frac{du}{u^2+1}=\frac{2\sqrt{7}}{7}\arctan \left(u\right)=$$$$=\frac{2\sqrt{7}}{2}\arctan \left(\frac{\sqrt{7}}{7}(2x+1)\right)$$$$$$$$=\frac{\sqrt{5}}{5}\ln \left(\frac{\mid 2x-\sqrt{5}-1\mid }{\mid 2x-\sqrt{5}+1\mid }\right)-\frac{2\sqrt{7}}{2}\arctan \left(\frac{\sqrt{7}}{7}(2x+1)\right)+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com