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Plz solve this integration..............
∫ ((√((x)^2+x+1))/(x+1)) dx

Sir in this,
x2+x+1 = (x+1)2 - x
where x+1 is replaced by t
then we get
((v((t)^2-(t-1)))/t) dt right; then how to proceed sir.....

where 'v' represents sq.rt


8 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans:$I = \int \frac{\sqrt{x^{2}+x+1}}{x+1}dx$$I = \int \frac{\sqrt{(x+\frac{1}{2})^{2}+\frac{3}{4}}}{x+1}dx$$u = x + \frac{1}{2}$$du = dx$$I = \int \frac{\sqrt{(u)^{2}+\frac{3}{4}}}{u+\frac{1}{2}}du$$u = \frac{\sqrt{3}}{2}tan(s)$$du = \frac{\sqrt{3}}{2}sec^{2}(s).ds$$I = \frac{3}{4}\int \frac{sec^{3}(s)}{\frac{\sqrt{3}}{2}tan(s)+\frac{1}{2}}ds$$t = tan (\frac{s}{2})$$dt = \frac{1}{2}sec^{2} (\frac{s}{2}).ds$$I = 3\int \frac{(p^{2}+1)^{2}}{(p^{2}-1)^{2}.(-p^{2}+2\sqrt{3}p+1)}dp$Use simply partial fraction rule here, you will get$I = \sqrt{x^{2}+x+1}-log(2 \sqrt{x^{2}+x+1}-(x-1))+log(x+1)-\frac{1}{2}sinh^{-1}(\frac{2x+1}{\sqrt{3}}) + c$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
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