If ∫ (t + 1 + t ^ 2 ) 1 + t ^ 2 dt = (1)/(2) (g(t)) ^ 2 + c , where c is a constant, then g(2) , is equal to

Correct answer: C. (2 + 5 ) Put, (t + 1 + t ^ 2 ) = z ⇒ 1 (t + 1 + t ^ 2 ) (1 + 2t 2 1 + t ^ 2 )dt = dz ⇒ (t + 1 + t ^ 2 ) (t + 1 + t ^ 2 ) 1 1 + t ^ 2 dt = dz ⇒ 1 1 + t ^ 2 dt = dz ∴ I = ∫ zdz = z ^ 2 2 + c = (1)/(2) [ (t + 1 + t ^ 2 )] ^ 2 + c , where c is the constant of integration. ∴ g(t) = (t + 1 + t ^ 2 ) ⇒ g(2) = (2 + 1 + 4 ) Hence, (2 + 5 ) , is the answer.

JEE Main 201511 AprilOnline

Exam: JEE Main
Year: 2015
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2015 · 11 April · Online · Mathematics · Indefinite Integration

If ∫ (t + 1 + t ^ 2 ) 1 + t ^ 2 dt = (1)/(2) (g(t)) ^ 2 + c , where...

∫ \dfrac{\log \left(t + \sqrt{1 + {t}^{2}}\right)}{\sqrt{1 + {t}^{2}}}dt = \dfrac{1}{2}{\left(g\left(t\right)\right)}^{2} + c

, where

c

is a constant, then

g\left(2\right)

, is equal to

Options

A $\dfrac{1}{\sqrt{5}}\log \left(2 + \sqrt{5}\right)$
B $2\log \left(2 + \sqrt{5}\right)$
C $\log \left(2 + \sqrt{5}\right)$
D $\dfrac{1}{2}\log ( 2 + \sqrt{5} )$

Correct answer

C. $\log \left(2 + \sqrt{5}\right)$

Step-by-step solution

Put,

\log \left(t + \sqrt{1 + {t}^{2}}\right) = z

⇒ \dfrac{1}{\left(t + \sqrt{1 + {t}^{2}}\right)}\left(1 + \dfrac{2t}{2\sqrt{1 + {t}^{2}}}\right)dt = dz

⇒ \dfrac{\left(t + \sqrt{1 + {t}^{2}}\right)}{\left(t + \sqrt{1 + {t}^{2}}\right)}\dfrac{1}{\sqrt{1 + {t}^{2}}}dt = dz

⇒ \dfrac{1}{\sqrt{1 + {t}^{2}}}dt = dz

∴ I = ∫ zdz = \dfrac{{z}^{2}}{2} + c = \dfrac{1}{2}{\left[\log \left(t + \sqrt{1 + {t}^{2}}\right)\right]}^{2} + c

, where

c

is the constant of integration.

∴ g\left(t\right) = \log \left(t + \sqrt{1 + {t}^{2}}\right)

⇒ g\left(2\right) = \log \left(2 + \sqrt{1 + 4}\right)

Hence,

\log \left(2 + \sqrt{5}\right)

, is the answer.

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