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Question Number 82531 by Power last updated on 22/Feb/20

Commented by abdomathmax last updated on 24/Feb/20

$$I=\int \frac{\sqrt{1-x}}{x}dx+\int \frac{\sqrt{2-x^2}}{x}dx+\int \frac{\sqrt{4-x^4}}{x}dxwehave$$$$\int \frac{\sqrt{1-x}}{x}dxchangement\sqrt{1-x}=tgive1-x=t^2\Rightarrow$$$$x=1-t^2\Rightarrow \int \frac{\sqrt{1-x}}{x}dx=\int \frac{t}{1-t^2}(-2t)dt$$$$=\int \frac{2t^2}{t^2-1}dt=2\int \frac{t^2-1+1}{t^2-1}dt=2\int dt+2\int \frac{dt}{t^2-1}$$$$=2t+\int (\frac{1}{t-1}-\frac{1}{t+1})dt=2t+ln\mid \frac{t-1}{t+1}\mid +c$$$$=2\sqrt{1-x}+ln\mid \frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\mid +c$$$$\int \frac{\sqrt{2-x^2}}{x}dx{=}_{x=\sqrt{2}sint}\int \frac{\sqrt{2}cost}{\sqrt{2}sint}\times \sqrt{2}costdt$$$$=\sqrt{2}\int \frac{{cos}^{2}t}{sint}dt=\sqrt{2}\int \frac{1-{sin}^{2}t}{sint}dt$$$$=\sqrt{2}\int \frac{dt}{sint}-\sqrt{2}\int sintdt$$$${=}_{tan\left(\frac{t}{2}\right)=u}\sqrt{2}\int \frac{1}{\frac{2u}{1+u^2}}\times \frac{2du}{1+u^2}+\sqrt{2}cost+c^{{}^{\prime }}$$$$=\sqrt{2}ln\mid tan\left(\frac{t}{2}\right)\mid +\sqrt{2}\sqrt{1-{\left(\frac{x}{\sqrt{2}}\right)}^2}+c^{{}^{\prime }}$$$$=\sqrt{2}ln\mid tan\left(\frac{1}{2}arcsin\left(\frac{x}{\sqrt{2}}\right)\right)\mid +\sqrt{2}\sqrt{1-\frac{x^2}{2}}+c$$$$....becontinued...$$

Answered by TANMAY PANACEA last updated on 22/Feb/20

$$\int \frac{1-x}{x\sqrt{1-x}}dx+\int \frac{2-x^2}{x\sqrt{2-x^2}}dx+\int \frac{4-x^4}{x\sqrt{4-x^4}}dx$$$$I_1=\int \frac{1-x}{x\sqrt{1-x^2}}dx$$$$t^2=1-x$$$$2tdt=-dx$$$$\int \frac{t^2}{(1-t^2)t}\times -2tdt$$$$2\int \frac{1-t^2-1}{1-t^2}dt$$$$2\int dt-2\int \frac{dt}{1-t^2}$$$$2\int dt-[\int \frac{dt}{1+t}+\int \frac{dt}{1-t}]$$$$2\int dt-\int \frac{dt}{1+t}+\int \frac{dt}{t-1}$$$$2t+ln\left(\frac{t-1}{t+1}\right)+c_1$$$$\sqrt{1-x}+ln\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)+c_1$$$$I_2=\int \frac{2-x^2}{x\sqrt{2-x^2}}$$$$2\int \frac{xdx}{x^2\sqrt{2-x^2}}-\int \frac{xdx}{\sqrt{2-x^2}}$$$$k^2=2-x^2\to kdk=-xdx$$$$2\int \frac{-kdk}{(2-k^2)}+\int \frac{kdk}{k}$$$$\int \frac{d(2-k^2)}{2-k^2}+\int dk$$$$ln(2-k^2)+k+c_2$$$$ln\left(x^2\right)+\sqrt{2-x^2}+c_2$$$$I_3=\int \frac{4-x^4}{x\sqrt{4-x^4}}dx$$$$\int \frac{4xdx}{x^2\sqrt{4-x^4}}-\int \frac{x^3}{\sqrt{4-x^4}}dx$$$$2\int \frac{d\left(x^2\right)}{x^2\sqrt{4-x^4}}-\frac{1}{2}\int \frac{x^2\times d\left(x^2\right)}{\sqrt{4-x^4}}$$$$2\int \frac{dp}{p\sqrt{4-p^2}}-\frac{1}{2}\int \frac{pdp}{\sqrt{4-p^2}}[p=x^2]$$$$2\int \frac{pdp}{p^2\sqrt{4-p^2}}-\frac{1}{2}\int \frac{pdp}{\sqrt{4-p^2}}$$$$a^2=4-p^2\to 2ada=-2pdp★[a^2=4-p^2=4-x^4]$$$$2\int \frac{-ada}{(4-a^2)a}-\frac{1}{2}\int \frac{-ada}{a}$$$$-2\int \frac{da}{(2+a)(2-a)}+\frac{1}{2}\int da$$$$\frac{-1}{2}\int \frac{2+a+2-a}{(2+a)(2-a)}da+\frac{1}{2}\int da$$$$\frac{-1}{2}\int \frac{da}{2-a}-\frac{1}{2}\int \frac{da}{2+a}+\frac{1}{2}\int da$$$$\frac{1}{2}\int \frac{da}{a-2}-\frac{1}{2}\int \frac{da}{a+2}+\frac{1}{2}\int da$$$$\frac{1}{2}ln\left(\frac{a-2}{a+2}\right)+\frac{a}{2}+c_3$$$$\frac{1}{2}ln\left(\frac{\sqrt{4-x^4}-2}{\sqrt{4-x^4}+2}\right)+\frac{\sqrt{4-x^4}}{2}+c_3$$$$$$$$$$$$$$$$$$

Commented by peter frank last updated on 22/Feb/20

$$thankyou$$

Commented by Power last updated on 22/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by TANMAY PANACEA last updated on 22/Feb/20

$$mostwelcomebothofyousir...$$

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