The value of -1-x-12-22 d x isA- log- x -2- x -12-2B- log- x-2-x-22-1- C- log- x-1-x-12-2- D- log x-2-x-12-1 -

The correct option is C log | x 1 + √( ( x 1 )^2+2) |We know that ∫1/√(x^2 +a^2)dx nbsp;=log | x + √(x^2 +a^2 ) | Here , instead of x we have x 1 and th ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 2

The value of ...

Question

The value of $\int \frac{1}{\sqrt{{(x - 1)}^{2} + {(\sqrt{2})}^{2}}} d x$ is

log ∣ ∣ ∣ x - 2 + \sqrt{{(x - 1)}^{2} + 2} ∣ ∣ ∣

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log ∣ ∣ ∣ x - 2 + \sqrt{{(x - 2)}^{2} + 1} ∣ ∣ ∣

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log ∣ ∣ ∣ x - 1 + \sqrt{{(x - 1)}^{2} + 2} ∣ ∣ ∣

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log ∣ ∣ ∣ x - 2 + \sqrt{{(x - 1)}^{2} + 1} ∣ ∣ ∣

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Solution

The correct option is C $log ∣ ∣ ∣ x - 1 + \sqrt{{(x - 1)}^{2} + 2} ∣ ∣ ∣$
We know that $\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = log ∣ ∣ x + \sqrt{x^{2} + a^{2}} ∣ ∣$
Here , instead of x we have x - 1 and the value of a will be = $\sqrt{2}$

So $\int \frac{1}{\sqrt{(x - 1)^{2} + (\sqrt{2})^{2}}} d x$ would be equal to $log | x - 1 + \sqrt{(x - 1)^{2} + 2} |$

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Similar questions

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

l o g | (x - 1) + q | + C

where Q=

Q. The inverse of the function

f : R \to \{x \in R : x < 1\}

given by

f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}

is
(a)

\frac{1}{2} \log \frac{1 + x}{1 - x}

(b)

\frac{1}{2} \log \frac{2 + x}{2 - x}

(c)

\frac{1}{2} \log \frac{1 - x}{1 + x}

(d) none of these

Q. If

f (\frac{x - 4}{x + 2}) = 2 x + 1, (x \in R - {1, - 2}),

then

\int f (x) d x

is equal to :
(where

C

is a constant of integration)

\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x =

(a)

\frac{π}{12} + \log (2 \sqrt{2})

(b)

\frac{π}{2} + \log (2 \sqrt{2})

(c)

\frac{π}{6} + \log (2 \sqrt{2})

(d)

\frac{π}{3} + \log (2 \sqrt{2})

Q. If

{log}_{y} x + {log}_{x} y = 1, x^{2} + y = 12

, then the value of xy is:

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The value of ∫1√(x−1)2+(√2)2dx is

The correct option is C log∣∣∣x−1+√(x−1)2+2∣∣∣We know that ∫1√x2+a2dx=log∣∣x+√x2+a2∣∣ Here , instead of x we have x - 1 and the value of a will be =√2 So ∫1√(x−1)2+(√2)2dx would be equal to log|x−1+√(x−1)2+2|

The value of $\int \frac{1}{\sqrt{{(x - 1)}^{2} + {(\sqrt{2})}^{2}}} d x$ is