If f ( x ) = √ x , g ( x ) = e x − 1 , a n d ∫ f o g ( x ) d x = A f o g ( x ) + B t a n − 1 ( f o g ( x ) ) + C , then A + B is equal to Options: A . &nbsp 1 B . &nbsp 2 C . &nbsp 3 D . &nbsp 0

Answer: Option D : D f o g ( x ) = √ e x − 1 ∴ I = ∫ √ e x − 1 d x = ∫ 2 t 2 t 2 + 1 d t { w h e r e t = √ e x − 1 } = 2 t − 2 t a n − 1 t + C = 2 √ e x − 1 − 2 t a n − 1 ( √ e x − 1 ) + C = 2 f o g ( x ) − 2 t a n − 1 ( f o g ( x ) ) + C ∴ A + B = 2 + ( − 2 ) = 0

If f(x)=√x,g(x)=ex−1,and∫fog(x)dx=Afog(x)+Btan−1(f

Discussion Forum : Indefinite Integration

Question -

If $f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C$ , then A + B is equal to

Options:

A . 1

B . 2

C . 3

D . 0

Answer: Option D
:
D

f o g (x) = \sqrt{e^{x} - 1} ∴ I = \int \sqrt{e^{x} - 1} d x = \int \frac{2 t^{2}}{t^{2} + 1} d t {w h e r e t = \sqrt{e^{x} - 1}} = 2 t - 2 t a n^{- 1} t + C = 2 \sqrt{e^{x} - 1} - 2 t a n^{- 1} (\sqrt{e^{x} - 1}) + C = 2 f o g (x) - 2 t a n^{- 1} (f o g (x)) + C ∴ A + B = 2 + (- 2) = 0

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Discussion Forum : Indefinite Integration

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