If ∫ ( x − 1 x + 1 ) d x √ x 3 + x 2 + x = 2 t a n − 1 √ f ( x ) + C , find f(x). Options: A . &nbsp x + 1 x + 1 B . &nbsp x + 1 x + 2 C . &nbsp x − 1 x + 1 D . &nbsp x − 1 x − 1

Answer: Option A : A I = ∫ ( x − 1 ) d x ( x + 1 ) x √ x + 1 + 1 x = ∫ ( x − 1 ) ( x + 1 ) d x ( x + 1 ) 2 x √ x + 1 + 1 x = ∫ ( 1 − 1 x 2 ) d x ( x + 1 x + 2 ) √ x + 1 x + 1 P u t x + 1 + 1 x = t 2 ( 1 − 1 x 2 ) d x = 2 t d t = ∫ 2 t d t ( t 2 + 1 ) t = 2 t a n − 1 t + c = 2 t a n − 1 ( √ x + 1 x + 1 ) + c f(x) = x + 1 x + 1

If ∫(x−1x+1)dx√x3+x2+x=2tan−1√f(x)+C, find

Discussion Forum : Indefinite Integration

Question -

If $\int (\frac{x - 1}{x + 1}) \frac{d x}{\sqrt{x^{3} + x^{2} + x}} = 2 t a n^{- 1} \sqrt{f (x)} + C$ , find f(x).

Options:

A .

x + \frac{1}{x} + 1

B .

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

C .

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

D .

x - \frac{1}{x} - 1

Answer: Option A
:
A

I = \int \frac{(x - 1) d x}{(x + 1) x \sqrt{x + 1 + \frac{1}{x}}} = \int \frac{(x - 1) (x + 1) d x}{(x + 1)^{2} x \sqrt{x + 1 + \frac{1}{x}}} = \int \frac{(1 - \frac{1}{x^{2}}) d x}{(x + \frac{1}{x} + 2) \sqrt{x + \frac{1}{x} + 1}} P u t x + 1 + \frac{1}{x} = t^{2} (1 - \frac{1}{x^{2}}) d x = 2 t d t = \int \frac{2 t d t}{(t^{2} + 1) t} = 2 t a n^{- 1} t + c = 2 t a n^{- 1} (\sqrt{x + \frac{1}{x} + 1}) + c

f(x) =

x + \frac{1}{x} + 1

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Discussion Forum : Indefinite Integration

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