-d x-x-a-x-b-A- None of theseB- 2-3 b - a - x - a 3 - 2- x - b 3 - 2- cC- 2-3a-b-x-a3 - 2-x-b3 - 2-cD- 2-3a-b-x-a3 - 2-x-b3 - 2-c

The correct option is D 2/3(a b)[(x+a)^3/2 (x+b)^3/2]+c∫dx/√(x+a)+√(x+b)=∫√(x+a) √(x+b)/(x+a) (x+b)dx =1/(a b)∫ (x+a)^1/2dx 1/(a b)∫ (x+b)^1/2dx =2/3(a b)[(x ...

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Byju's Answer

Standard XII

Physics

Definite Integrals

∫d x/√x+a+√x+...

Question

$\int \frac{d x}{\sqrt{x + a} + \sqrt{x + b}} =$

\frac{2}{3 (b - a)} [(x + a)^{3 / 2} - (x + b)^{3 / 2}] + c

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\frac{2}{3 (a - b)} [(x + a)^{3 / 2} - (x + b)^{3 / 2}] + c

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\frac{2}{3 (a - b)} [(x + a)^{3 / 2} + (x + b)^{3 / 2}] + c

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None of these

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Solution

The correct option is B $\frac{2}{3 (a - b)} [(x + a)^{3 / 2} - (x + b)^{3 / 2}] + c$
$\int \frac{d x}{\sqrt{x + a} + \sqrt{x + b}} = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} d x$
$= \frac{1}{(a - b)} \int (x + a)^{1 / 2} d x - \frac{1}{(a - b)} \int (x + b)^{1 / 2} d x$
$= \frac{2}{3 (a - b)} [(x + a)^{3 / 2} - (x + b)^{3 / 2}] + c$

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Similar questions

Q. Get the factorisation

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

writing the expression

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = [{(a + b + c)}^{3} - a^{3}] - [b^{3} + c^{3}]

\frac{(a + b)^{3} + (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a)}{3 (a^{3} + b^{3} + c^{3} - 3 a b c)}

= ____________

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a^{3} + b^{3} = (8 - 3 a - 3 b) a b

and show that

log \frac{a + b}{2} = \frac{1}{3} (log a + log b)

Q. If 3x = a + b + c, then the value of (x − a)³ + (x −b)³ + (x − c)³ − 3(x − a) (x − b) (x −c) is

(a) a + b + c

(b) (a − b) (b − c) (c − a)

(c) 0

(d) none of these

Q. Prove:

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

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∫dx√x+a+√x+b=

The correct option is B 23(a−b)[(x+a)3/2−(x+b)3/2]+c∫dx√x+a+√x+b=∫√x+a−√x+b(x+a)−(x+b)dx =1(a−b)∫(x+a)1/2dx−1(a−b)∫(x+b)1/2dx =23(a−b)[(x+a)3/2−(x+b)3/2]+c

$\int \frac{d x}{\sqrt{x + a} + \sqrt{x + b}} =$