The value of ∫ xdx ( x + 3 ) x + 1 is (where, c & is the constant of integration)

Correct answer: C. 2 x + 1 - 3 2 tan^ - 1 (x + 1)/(2) + c Substitute& x + 1 = t , we get, 2 ∫ t^ 2 - 1 t^ 2 + 2 dt = 2t - 6 ∫ 1 t^ 2 + 2 dt = 2t - 3 2 tan^ - 1 t 2 + c = 2 x + 1 - 3 2 tan^ - 1 (x + 1)/(2) + c

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of ∫ xdx ( x + 3 ) x + 1 is (where, c & is the constant...

The value of

∫ \dfrac{xdx}{( x + 3 ) \sqrt{x + 1}}

is (where,

c

& is the constant of integration)

Options

A $2\sqrt{x + 1} + 3tan^{- 1}\sqrt{x + 1} + c$
B $2\sqrt{x + 1} + 3\sqrt{2}tan^{- 1}\sqrt{\dfrac{x + 1}{2}} + c$
C $2\sqrt{x + 1} - 3\sqrt{2}tan^{- 1}\sqrt{\dfrac{x + 1}{2}} + c$
D $2\sqrt{x + 1} - 3tan^{- 1}\sqrt{x + 1} + c$

Correct answer

C. $2\sqrt{x + 1} - 3\sqrt{2}tan^{- 1}\sqrt{\dfrac{x + 1}{2}} + c$

Step-by-step solution

Substitute& $\sqrt{x + 1} = t$ , we get,

$2 ∫ \dfrac{t^{2} - 1}{t^{2} + 2}dt = 2t - 6 ∫ \dfrac{1}{t^{2} + 2}dt$

$= 2t - 3\sqrt{2}tan^{- 1}\dfrac{t}{\sqrt{2}} + c$

$= 2\sqrt{x + 1} - 3\sqrt{2}tan^{- 1}\sqrt{\dfrac{x + 1}{2}} + c$

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