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Question Number 26397 by abdo imad last updated on 25/Dec/17

$$find\int \frac{dx}{x\sqrt{1+x^2}}andcalculate{\int }_1^3\frac{dx}{x\sqrt{1+x^2}}$$

Commented by abdo imad last updated on 26/Dec/17

$$letusethechangementx=tan\theta$$$$\int \frac{dx}{x\sqrt{1+x^2}}=\int \frac{(1+{tan}^2\theta )d\theta }{tan\theta \sqrt{1+{tan}^2}}d\theta =\int \frac{\sqrt{1+{tan}^2\theta }}{tan\theta }d\theta$$$$=\int \frac{1}{cos\theta tan\theta }d\theta =\int \frac{d\theta }{sin\theta }andbythechangementtan\left(\theta /2\right)=t$$$$I=\int \frac{2dt}{1+t^2}{\left(\frac{2t}{1+t^2}\right)}^{-1}=\int dt/t=ln/t/=ln/tan\left(\frac{arctanx}{2}\right)/ln$$$$\Rightarrow \int \frac{dx}{x\sqrt{1+x^2}}=ln/tan\left(\frac{arctanx}{2}\right)/+k$$$${\int }_1^3\frac{dx}{x\sqrt{1+x^2}}=ln/tan\left(2^{-1}arctan3\right)/-ln/tan\left(2^{-1}arctan1\right)/$$$$==ln/tan\left(\frac{arctan3}{2}\right)/-ln/tan\left(\frac{\pi }{8}\right)/.$$

Answered by $@ty@m last updated on 25/Dec/17

$$x=\tan y$$$$dx=\sec {}^{2}ydy$$$$I=\int \frac{\sec {}^{2}ydy}{\tan y.\sec y}$$$$=\int \mathrm{cosec}ydy$$$$=\ln \mid \mathrm{cosec}y-\cot y\mid +C$$$$=\ln \mid \mathrm{cosec}\left({\tan }^{-1}x\right)-\cot \left({\tan }^{-1}x\right)\mid +C$$

Answered by $@ty@m last updated on 25/Dec/17

$$Anothermethod:$$$$Letx=\frac{1}{t}$$$$dx=-\frac{1}{t^2}dt$$$$\int \frac{-\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{1+\frac{1}{t^2}}}$$$$\int \frac{-\frac{1}{t^2}dt}{\frac{1}{t^2}\sqrt{t^2+1}}$$$$=\int \frac{-dt}{\sqrt{1+t^2}}$$$$=-\ln \mid t+\sqrt{1+t^2}\mid +C$$$$=-\ln \mid \frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\mid +C$$$$=-\ln \mid \frac{1+\sqrt{1+x^2}}{x}\mid +C$$$$=\ln \mid \frac{x}{1+\sqrt{1+x^2}}\mid +C$$

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