Let ∫ dx x^ 2 + 1 - x = f(x) + C such that f (0) = 0 and C is the constant of integration, then the value of f(1) is

Correct answer: D. 1 2 +(1)/(2) (1 + (1 + 2 )) ∫ ( x^ 2 + 1 + x) 1 dx ∫ ( x^ 2 + 1 )dx + ∫ xdx = (x)/(2) x^ 2 + 1 + (1)/(2) |(x + x^ 2 + 1 )| + x^ 2 2 + C f(x) = (x)/(2) x^ 2 + 1 + (1)/(2) |x + 1 x^ 2 + 1 | + (x2)/(2) f(0) = 0 f(1) = 2 2 + (1)/(2) (1 + 2 ) + (1)/(2) = 1 2 + (1)/(2) (1 + (1 + 2 ))

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let ∫ dx x^ 2 + 1 - x = f(x) + C such that f (0) = 0 and C is the...

Let

∫ \dfrac{dx}{\sqrt{x^{2} + 1} - x} = f\left(x\right) + C

such that

f \left(0\right) = 0

and

C

is the constant of integration, then the value of

f\left(1\right) is

Options

A $\dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\ln \left(1 +$ \sqrt{2}\right)
B $\dfrac{1}{2} + \dfrac{1}{\sqrt{2}}\ln \left(1 +$ \sqrt{2}\right)
C $\dfrac{1}{2} + \dfrac{1}{2}ln \sqrt{2} + 1$
D $\dfrac{1}{\sqrt{2}} +\dfrac{1}{2} \left(1 + \ln \left(1 +$ \sqrt{2}\right)\right)

Correct answer

D. $\dfrac{1}{\sqrt{2}} +\dfrac{1}{2} \left(1 + \ln \left(1 +$ \sqrt{2}\right)\right)

Step-by-step solution

∫ \dfrac{\left(\sqrt{x^{2} + 1} + x\right)}{1}dx

∫ \left(\sqrt{x^{2} + 1}\right)dx + ∫ xdx

= \dfrac{x}{2}\sqrt{x^{2} + 1} + \dfrac{1}{2}\ln \left|\left(x + \sqrt{x^{2} + 1}\right)\right| + \dfrac{x^{2}}{2} + C

f\left(x\right) = \dfrac{x}{2}\sqrt{x^{2} + 1} + \dfrac{1}{2}\ln \left|x + 1\sqrt{x^{2} + 1}\right| + \dfrac{x2}{2}

f\left(0\right) = 0

f\left(1\right) = \dfrac{\sqrt{2}}{2} + \dfrac{1}{2}\ln \left(1 + \sqrt{2}\right) + \dfrac{1}{2} = \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \left(1 + \ln \left(1 +

\sqrt{2}\right)\right)

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