# Integral doubt

Note by Rishabh Bhatnagar
1 year, 7 months ago

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2) Use the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ $I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{x}+\sqrt{2012-x}}dx$ Therefore,$2I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx+\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}dx$ Therefore, $I=1005$

- 1 year, 7 months ago

1) By algebric manupilations the given integral can be converted into -$\int \frac{(1-\frac{1}{x^{2}})}{(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}dx$ Now by u-substitution method-$u^{2}=x+\frac{1}{x}+1$ And $(1-\frac{1}{x^{2}})dx=2u\cdot du$ Now the problem can be solved using standard integrals.

- 1 year, 7 months ago

thanx a lot

- 1 year, 7 months ago