- x 2-2 x -5 dx is equal toA- -frac12x-1- sqrt x- 2-2 x-5-2 log-x-1-sqrtx- 2-2 x-5-C lB- x-1 sqrtx- 2-2 x-5-2 log -x-1- sqrtx- 2-2 x-5-C-C- x-1- isqrt x- 2-2 x-5 -2 log- x -1-lsqrt x - 2-2 x -5- Cl D- x-1- isqrt x- 2-2 x-5-frac12log- x -1-sqrt x - 2-2 x -5- Cl

The correct option is A 1/2(x+1) √((x^2+2x+5)) +2log| x+1+ √((x^2+2x+5))|+C∫√(x^2+a^2)dx = x(√(x^2+a^2))/2+a^2/2log|x+√(x^2+a^2)| ∫√((x^2+2x+5)) dx = ∫√((x+1) ...

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Byju's Answer

Standard XII

Mathematics

Quotient Rule of Differentiation

∫√ x 2+2 x +5...

Question

$\int \sqrt{(x^{2} + 2 x + 5)} d x$ is equal to

\frac{1}{2} (x + 1) \sqrt{(x^{2} + 2 x + 5)} + 2 log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C

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(x + 1) \sqrt{(x^{2} + 2 x + 5)} + 2 log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C

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(x + 1) \sqrt{(x^{2} + 2 x + 5)} - 2 log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C

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(x + 1) \sqrt{(x^{2} + 2 x + 5)} + \frac{1}{2} log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C

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Solution

The correct option is A $\frac{1}{2} (x + 1) \sqrt{(x^{2} + 2 x + 5)} + 2 log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C$
$\int \sqrt{x^{2} + a^{2}} d x = \frac{x (\sqrt{x^{2} + a^{2}})}{2} + \frac{a^{2}}{2} log | x + \sqrt{x^{2} + a^{2}} |$
$\int \sqrt{(x^{2} + 2 x + 5)} d x = \int \sqrt{(x + 1)^{2} + 2^{2}} d x$
$∴ \int \sqrt{(x^{2} + 2 x + 5)} d x = \frac{1}{2} (x + 1) \sqrt{(x^{2} + 2 x + 5)} + 2 log | x + 1 + \sqrt{(x^{2} + 2 x + 5)} | + C$

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Similar questions

\int \sqrt{x^{2} + 2 x + 5} dx is equal to

Q. Solve for

x

2^{{log}_{\sqrt{2}} (x - 1)} > x + 5

Evaluate $\int \frac{1}{(1 - x^{2}) \sqrt{1 + x^{2}}} d x$

Q. Prove that

\int \frac{x + 3}{\sqrt{(x^{2} + 2 x + 2)}} d x = \sqrt{x^{2} + 2 x + 2} + 2 log [x + 1 - {(x + 1)^{2} + 1}^{1 / 2}] .

I = \int \frac{x^{3}}{\sqrt{x^{2} + 2 x + 5}} d x = \frac{1}{6} f (x) - \frac{A}{120} g (x) + C

where

f (x) = (2 x^{2} - 5 x - 36) \sqrt{x^{2} + 2 x + 5}, g (x) = log (x + 1 + \sqrt{x^{2} + 2 x + 5})

. Then, A is equal to

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∫√(x2+2x+5) dx is equal to

The correct option is A 12(x+1)√(x2+2x+5)+2log|x+1+√(x2+2x+5)|+C∫√x2+a2dx = x(√x2+a2)2+a22log|x+√x2+a2| ∫√(x2+2x+5) dx = ∫√(x+1)2+22 dx ∴∫√(x2+2x+5) dx = 12(x+1)√(x2+2x+5) +2log|x+1+√(x2+2x+5)|+C

$\int \sqrt{(x^{2} + 2 x + 5)} d x$ is equal to