Integral doubt

hello, i am stuck in these integrals can anyone please help

#Calculus

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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, $I=1005$

Aaron Jerry Ninan - 4 years, 4 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.

Aaron Jerry Ninan - 4 years, 4 months ago

thanx a lot

Rishabh Bhatnagar - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$