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Question Number 37004 by behi83417@gmail.com last updated on 07/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

$$\int \frac{\sqrt{1+x^p}}{x}dx$$$$\int \frac{1+x^p}{x\sqrt{1+x^p}}dx$$$$\int \frac{x^{p-1}(1+x^p)}{x^p\sqrt{1+x^p}}dx$$$$t=x^p$$$$dt={px}^{p-1}dx$$$$\int \frac{dt}{p}\times \frac{(1+t)}{t\sqrt{1+t}}$$$$\frac{1}{p}\int \frac{dt}{t\sqrt{1+t}}+\frac{1}{p}\int \frac{dt}{\sqrt{1+t}}$$$$k^2=1+t2kdk=dt$$$$\frac{1}{p}\int \frac{2kdk}{(k^2-1)k}+\frac{1}{p}\int \frac{2kdk}{k}$$$$\frac{2}{p}\int \frac{dk}{k^2-1}+\frac{2}{p}\int dk$$$$\frac{2}{p}\times \frac{1}{2}\int \frac{(k+1)-(k-1)}{(k+1)(k-1)}dk+\frac{2}{p}\int dk$$$$\frac{1}{p}ln\mid \frac{k-1}{k+1}\mid +\frac{2}{p}k+c$$$$\frac{1}{p}ln\mid \frac{\sqrt{1+x^p}-1}{\sqrt{1+x^p}+1}\mid +\frac{2}{p}k$$$$\frac{1}{p}ln\mid \frac{\sqrt{1+x^p}-1}{\sqrt{1+x_{}^p}+1}\mid +\frac{2}{p}\sqrt{1+x^p}$$

Commented by behi83417@gmail.com last updated on 07/Jun/18

$$mrtanmay!niceworkdonebyyou.$$$$thanks!$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

$$itsalright...thankyousir$$

Answered by ajfour last updated on 07/Jun/18

$$let{x}^p=\tan {}^2\theta$$$$\Rightarrow {px}^{p-1}dx=2\tan \theta \sec {}^2\theta d\theta$$$$\Rightarrow {px}^p\left(\frac{dx}{x}\right)=2\tan \theta \sec {}^2\theta d\theta$$$$orp\tan {}^2\theta \left(\frac{dx}{x}\right)=2\tan \theta \sec {}^2\theta d\theta$$$$I=\int \frac{\sec \theta \left(2\tan \theta \sec {}^2\theta \right)d\theta }{p\tan {}^2\theta }$$$$=\frac{2}{p}\int \frac{d\theta }{\sin \theta \cos {}^2\theta }$$$$=\frac{2}{p}\int \frac{\sin \theta d\theta }{\cos {}^2\theta (1-\cos {}^2\theta )}$$$$=\frac{2}{p}\int \frac{dt}{t^2(t^2-1)}(t=\cos \theta )$$$$\frac{1}{t^2(t-1)(t+1)}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t-1}+\frac{D}{t+1}$$$$C=\frac{1}{2},D=-\frac{1}{2},B=-1$$$$coeff.of{t}^{3}isA+C+D=0$$$$\Rightarrow A=0$$$$\therefore I=\frac{2}{p}[\frac{1}{t}+\frac{1}{2}\ln \mid \frac{t-1}{t+1}\mid ]+c$$$$As{x}^p=\tan {}^2\theta =\sec {}^2\theta -1=\frac{1}{t^2}-1$$$$\Rightarrow t=\frac{1}{\sqrt{1+x^p}}$$$$I=\frac{2}{p}[\sqrt{1+x^p}+\frac{1}{2}\ln \mid \frac{1-\sqrt{1+x^p}}{1+\sqrt{1+x^p}}\mid ]+c.$$$$forp=1$$$$I_1=2\sqrt{1+x^p}+\ln \mid \frac{1-\sqrt{1+x}}{1+\sqrt{1+x}}\mid +c.$$

Commented by behi83417@gmail.com last updated on 07/Jun/18

$$thanksinadvancesirAjfour!$$$$perfectmethod.$$

Answered by MJS last updated on 07/Jun/18

$$p=0$$$$\sqrt{2}\int \frac{1}{x}dx=\sqrt{2}\ln x+C$$$$p\ne 0$$$$\int \frac{\sqrt{1+x^p}}{x}dx=$$$$[t=x^p\to dx=\frac{x^{1-p}}{p}dt]$$$$=\frac{1}{p}\int \frac{\sqrt{1+t}}{t}dt=$$$$[u=\sqrt{1+t}\to dt=2\sqrt{1+t}du]$$$$=\frac{2}{p}\int \frac{u^2}{u^2-1}du=\frac{2}{p}(\int du+\int \frac{du}{u^2-1})=$$$$=\frac{2}{p}(u+\frac{1}{2}(\int \frac{du}{u-1}-\int \frac{du}{u+1}))=$$$$=\frac{2}{p}(u+\frac{1}{2}(\ln (u-1)-\ln (u+1)))=$$$$=\frac{2}{p}(u+\frac{1}{2}\ln \frac{u-1}{u+1})=\frac{2}{p}\sqrt{1+t}+\frac{1}{p}\ln \frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}=$$$$=\frac{2}{p}\sqrt{1+x^p}+\frac{1}{p}\ln \frac{\sqrt{1+x^p}-1}{\sqrt{1+x^p}+1}+C$$

Commented by behi83417@gmail.com last updated on 07/Jun/18

$$thankyouverymuchsir!$$$$simpleandsmart.$$

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