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Question Number 46720 by Necxx last updated on 30/Oct/18

Commented by Necxx last updated on 30/Oct/18

$$pleasehelpwithno.15cosigotit$$$$as$$$$-{\sin }^{-1}\left(\frac{1}{x}\right)-\frac{\sqrt{x^2+1}}{x}+c$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$letA=\int \frac{1}{x^2}\sqrt{\frac{x-1}{x+1}}dxchangement\frac{x-1}{x+1}=t^{2}givex-1=t^{2}x+t^2\Rightarrow$$$$(1-t^2)x=1+t^2\Rightarrow x=\frac{1+t^2}{1-t^2}\Rightarrow dx=\frac{2t(1-t^2)+2t(1+t^2)}{{(1-t^2)}^2}dt$$$$=\frac{2t-2t^3+2t+2t^3}{{(1-t^2)}^2}dt=\frac{4t}{{(1-t^2)}^2}dt\Rightarrow A=\int \frac{{(1-t^2)}^2}{{(1+t^2)}^2}t\frac{4t}{{(1-t^2)}^2}dt$$$$=4\int \frac{t^2}{{(1+t^2)}^2}dt=4\int \frac{1+t^2-1}{{(1+t^2)}^2}dt=4\int \frac{dt}{1+t^2}-4\int \frac{dt}{{(1+t^2)}^2}$$$$=4arctan\left(t\right)-4\int \frac{dt}{{(1+t^2)}^2}changementt=tan\theta give$$$$\int \frac{dt}{{(1+t^2)}^2}=\int \frac{1+{tan}^2\theta }{{(1+{tan}^2\theta )}^2}d\theta =\int \frac{d\theta }{1+{tan}^2\theta }=\int {cos}^2\theta d\theta =$$$$=\frac{1}{2}\int (1+cos\left(2\theta \right))d\theta =\frac{\theta }{2}+\frac{1}{4}sin\left(2\theta \right)=\frac{arctan\left(t\right)}{2}+\frac{1}{4}sin\left(2arctant\right)but$$$$sin\left(2arctant\right)=2sin\left(arctant\right)cos\left(arctant\right)$$$$=2\frac{t}{\sqrt{1+t^2}}.\frac{1}{\sqrt{1+t^2}}=\frac{2t}{1+t^2}\Rightarrow \int \frac{dt}{{(1+t^2)}^2}=\frac{arctan\left(t\right)}{2}+\frac{t}{2(1+t^2)}\Rightarrow$$$$A=2arctan\left(t\right)-\frac{2t}{1+t^2}+c$$$$=2arctan\left(\sqrt{\frac{x-1}{x+1}}\right)-\frac{2\sqrt{\frac{x-1}{x+1}}}{1+\frac{x-1}{x+1}}+c$$$$=2arctan\left(\sqrt{\frac{x-1}{x+1}}\right)-2\sqrt{\frac{x-1}{x+1}}\frac{x+1}{2x}+c$$$$A=2arctan\left(\sqrt{\frac{x-1}{x+1}}\right)+\frac{\sqrt{x^2-1}}{x}+c$$$$$$$$$$$$$$$$$$

Commented by Necxx last updated on 30/Oct/18

$$thankyousomuchsir.$$$$Theissuenowisthatthisanswer$$$$isnotintheoption.Doesitmean$$$$theoptionsareallwrong?$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$yesthereissomeerrorinthegivenanswersbutalwaysyoumusttrust$$$$yourselfandyourmethod...$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$16)letI=\int \frac{dx}{\sqrt{x-x^2}}\Rightarrow I=\int \frac{dx}{\sqrt{-(x^2-x)}}=\int \frac{dx}{\sqrt{-(x^2-2\frac{1}{2}x+\frac{1}{4}-\frac{1}{4})}}$$$$=\int \frac{dx}{\sqrt{\frac{1}{4}-{(x-\frac{1}{2})}^2}}{=}_{x-\frac{1}{2}=\frac{1}{2}sint}\int \frac{1}{2.\frac{1}{2}cost}costdt$$$$=\int dt=t+c=arcsin(2x-1)+csooption\left(B\right)istheanswer.$$

Commented by Necxx last updated on 30/Oct/18

$$wow....isolvedthisandgotAas$$$$myanswer$$$$$$

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