Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 86062 by ar247 last updated on 26/Mar/20

$$\int \frac{\sqrt{x^2-25}}{x}dx$$

Commented by abdomathmax last updated on 27/Mar/20

$$I=\int \frac{\sqrt{x^2-25}}{x}dxchangement\sqrt{x^2-25}=tgive$$$$x^2=t^2+25\Rightarrow xdx=tdt\Rightarrow dx=\frac{tdt}{x}\Rightarrow \frac{dx}{x}=\frac{tdt}{x^2}$$$$=\frac{tdt}{t^2+25}\Rightarrow I=\int t\left(\frac{tdt}{t^2+25}\right)$$$$=\int \frac{t^{2}dt}{t^2+25}=\int dt-\int \frac{25dt}{t^2+25}=t-25\int \frac{dt}{t^2+25}$$$${=}_{t=5u}t-25\int \frac{5du}{25(u^2+1)}=t-5arctan\left(\frac{t}{5}\right)+C$$$$=\sqrt{x^2-25}-5arctan\left(\frac{\sqrt{x^2-25}}{5}\right)+C$$

Answered by john santu last updated on 27/Mar/20

$$\sqrt{x^2-25}=t$$$$x^2=t^2+25\Rightarrow xdx=tdt$$$$\frac{dx}{x}=\frac{tdt}{t^2+25}$$$$\Rightarrow \int \frac{t.t}{t^2+25}dt=\int \frac{t^2+25-25}{t^2+25}dt$$$$=\int dt-25\int \frac{dt}{t^2+25}$$$$=t-{5\tan }^{-1}\left(\frac{t}{5}\right)+c$$$$=\sqrt{x^2-25}-{5\tan }^{-1}\left(\frac{\sqrt{x^2+25}}{5}\right)+c$$

Commented by john santu last updated on 27/Mar/20

$$oyes.thankyou$$$$$$

Commented by ar247 last updated on 26/Mar/20

$$$$$$$$$$$$$$$$$$\int dt-25\int \frac{dt}{t^2+25}=t-5{tan}^{-1}\left(\frac{t}{5}\right)+c$$$$right?$$$$$$$$$$

Commented by jagoll last updated on 26/Mar/20

$$\int 25\left(\frac{\mathrm{dt}}{{\mathrm{t}}^2+25}\right)=$$$$\mathrm{t}=5\tan \mathrm{u}\Rightarrow \mathrm{dt}=5\sec {}^2\mathrm{u}\mathrm{du}$$$$25\int \frac{5\sec {}^2\mathrm{u}\mathrm{du}}{25\sec {}^2\mathrm{u}}=5{\tan }^{-1}\left(\frac{\mathrm{u}}{5}\right)$$

Commented by jagoll last updated on 26/Mar/20

$$\mathrm{sir}\mathrm{john},\mathrm{typo}\mathrm{sir}\mathrm{in}\mathrm{line}6$$

Answered by Rio Michael last updated on 26/Mar/20

$$\int \frac{\sqrt{x^2-25}}{x}dx$$$$\mathrm{let}x=\frac{5}{\cos \theta }\Rightarrow \frac{\sqrt{x^2-25}}{x}=\sin x$$$$\int \frac{\sqrt{x^2-25}}{x}dx=5\int \frac{{\sin }^2\theta }{{\cos }^2\theta }d\theta =5\int {\tan }^2\theta d\theta$$$$=5(\tan \theta -\theta +k)$$$$\Rightarrow \int \frac{\sqrt{x^2-25}}{x}dx=5[\tan \left(\mathrm{arcos}\left(\frac{5}{x}\right)\right)-\mathrm{arcos}\left(\frac{5}{x}\right)+k]$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com