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Byju's Answer

Standard XII

Mathematics

Differentiation of a Determinant

If ∫dx1+√x+1+...

Question

If $\int \frac{d x}{1 + \sqrt{x + 1} + \sqrt{x}} = a x + b \sqrt{x} + c \int \sqrt{\frac{x + 1}{x}} d x$ upto constant of integration, where $a, b, c$ are constants, then the value of $(a + b + c)$ is

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Solution

$I = \int \frac{d x}{1 + \sqrt{x + 1} + \sqrt{x}}$
$= \int \frac{(1 + \sqrt{x} - \sqrt{x + 1})}{({(1 + \sqrt{x})}^{2} - 1 + x)} d x$
$= \int \frac{1 + \sqrt{x} - \sqrt{x + 1}}{2 \sqrt{x}} d x$
$= \frac{1}{2} \int x^{- 1 / 2} d x + \int \frac{d x}{2} - \frac{1}{2} \int \sqrt{\frac{x + 1}{x}} d x$
$= \frac{x}{2} + \sqrt{x} - \frac{1}{2} \int \sqrt{\frac{x + 1}{x}} d x$
$∴ a = \frac{1}{2}, b = 1, c = - \frac{1}{2}$
Hence, the value of $a + b + c$ is $1$

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If ∫dx1+√x+1+√x=ax+b√x+c∫√x+1xdx upto constant of integration, where a,b,c are constants, then the value of (a+b+c) is

I=∫dx1+√x+1+√x =∫(1+√x−√x+1)((1+√x)2−1+x)dx =∫1+√x−√x+12√xdx =12∫x−1/2dx+∫dx2−12∫√x+1xdx =x2+√x−12∫√x+1xdx ∴a=12,b=1,c=−12 Hence, the value of a+b+c is 1

If $\int \frac{d x}{1 + \sqrt{x + 1} + \sqrt{x}} = a x + b \sqrt{x} + c \int \sqrt{\frac{x + 1}{x}} d x$ upto constant of integration, where $a, b, c$ are constants, then the value of $(a + b + c)$ is