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 112251 by bobhans last updated on 07/Sep/20

$$\int \frac{\mathrm{dx}}{({\mathrm{x}}^4-1)\sqrt{{\mathrm{x}}^2+1}}$$

Answered by john santu last updated on 07/Sep/20

$$\int \frac{dx}{(x^4-1)\sqrt{x^2+1}}?$$$$setting:\tan r=x\Rightarrow dx=\sec {}^{2}rdr$$$$I=\int \frac{dx}{(x^2-1)(x^2+1)\sqrt{x^2+1}}$$$$I=\int \frac{\sec {}^{2}rdr}{(\tan {}^{2}r-1)\sec {}^{3}r}$$$$I=\int \frac{\cos rdr}{\tan {}^{2}r-1}=\int \frac{\cos {}^{3}rdr}{\sin {}^{2}r-\cos {}^{2}r}$$$$I=\int \frac{(1-\sin {}^{2}r)\cos rdr}{2\sin {}^{2}r-1}$$$$let\sin r=q\Rightarrow I=\int \frac{1-q^2}{2q^2-1}dq$$$$I=\frac{1}{2}\int (\frac{1}{2q^2-1}-1)dq$$$$I=\frac{1}{4}\int (-2+\frac{1}{q\sqrt{2}-1}-\frac{1}{q\sqrt{2}+1})dq$$$$I=\frac{1}{4}(-2q+\frac{1}{\sqrt{2}}\ln \mid q\sqrt{2}-1\mid -\frac{1}{\sqrt{2}}\ln \mid q\sqrt{2}+1\mid )+c$$$$I=\frac{1}{4}(\frac{-2x}{\sqrt{x^2+1}}+\frac{1}{\sqrt{2}}\ln \mid \frac{x\sqrt{2}-\sqrt{x^2+1}}{x\sqrt{2}+\sqrt{x^2+1}}\mid )+c$$

Commented by bemath last updated on 07/Sep/20

$$santuyyy$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com