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 168459 by BegamovSirojiddin last updated on 11/Apr/22

Answered by Mathspace last updated on 11/Apr/22

$$I=\int \frac{dx}{{(x^2+x+1)}^{\frac{5}{2}}}$$$$I=\int \frac{dx}{{({(x+\frac{1}{2})}^2+\frac{3}{4})}^{\frac{5}{2}}}(x+\frac{1}{2}=\frac{\sqrt{3}}{2}tant$$$$I=\int \frac{\frac{\sqrt{3}}{2}(1+{tan}^{2}t)dt}{{\left(\frac{3}{4}\right)}^{\frac{5}{2}}{(1+{tan}^{2}t)}^{\frac{5}{2}}}$$$$=\frac{\sqrt{3}}{2}{\left(\frac{4}{3}\right)}^{\frac{5}{2}}\int \frac{dt}{{(1+{tan}^{2}t)}^{\frac{3}{2}}}$$$$=\frac{\sqrt{3}}{2}{\left(\frac{4}{3}\right)}^{\frac{5}{2}}\int {cos}^{3}tdt$$$${cos}^{3}x={\left(\frac{e^{ix}+e^{-ix}}{2}\right)}^3$$$$=\frac{1}{8}(e^{3ix}+3e^{2ix}.e^{-ix}+3e^{ix}.e^{-2ix}+e^{-3ix})$$$$=\frac{1}{8}(e^{3ix}+e^{-3ix}+3(e^{ix}+e^{-ix})\}$$$$=\frac{1}{8}(2cos\left(3x\right)+6cosx)$$$$=\frac{1}{4}cos\left(3x\right)+\frac{3}{4}cosx\Rightarrow$$$$\int {cos}^{3}t=\frac{1}{4}\int cos\left(3t\right)dt+\frac{3}{4}\int costdt$$$$=\frac{1}{12}sin\left(3t\right)+\frac{3}{4}sint+\lambda$$$$but=t=arctan\left(\frac{2x+1}{\sqrt{3}}\right)\Rightarrow$$$$\Rightarrow I=\frac{\sqrt{3}}{2}{\left(\frac{4}{3}\right)}^{\frac{5}{2}}\{\frac{1}{12}sin(3arctan\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{3}{4}sin\left(arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)+\lambda \}$$

Answered by MJS_new last updated on 11/Apr/22

$$\int \frac{dx}{{(x^2+x+1)}^{5/2}}=$$$$[t=\frac{2x+1+2\sqrt{x^2+x+1}}{\sqrt{3}}\to dx=\frac{\sqrt{x^2+x+1}}{t}dt]$$$$=\frac{256}{9}\int \frac{t^3}{{(t^2+1)}^4}dt=\frac{256}{9}\int (\frac{t}{{(t^2+1)}^3}-\frac{t}{{(t^2+1)}^4})dt=$$$$=-\frac{64}{9{(t^2+1)}^2}+\frac{128}{27{(t^2+1)}^3}=-\frac{64(3t^2+1)}{27{(t^2+1)}^3}=$$$$=\frac{2(2x+1)(8x^2+8x+11)}{27{(x^2+x+1)}^{3/2}}+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com