If -x-1-x-1 d x-x3-x2-x-2 tan -1-fx-C- find fx

The correct option is A x + 1/x + 1I= ∫(x 1)dx/(x+1)x √(x + 1 + 1/x) = ∫(x 1) (x+1)dx/(x+1)^2x √(x + 1 + 1/x) = ∫(1 1/x^2)dx/(x + 1/x + 2)√(x + 1/x + 1) Put ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If ∫x-1/x+1 d...

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).

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

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

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

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

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Solution

The correct option is A $x + \frac{1}{x} + 1$
$I = \int \frac{(x - 1) d x}{(x + 1) x \sqrt{x + 1 + \frac{1}{x}}} = \int \frac{(x - 1) (x + 1) d x}{(x + 1)^{2} x \sqrt{x + 1 + \frac{1}{x}}} = \int \frac{(1 - \frac{1}{x^{2}}) d x}{(x + \frac{1}{x} + 2) \sqrt{x + \frac{1}{x} + 1}} P u t x + 1 + \frac{1}{x} = t^{2} (1 - \frac{1}{x^{2}}) d x = 2 t d t = \int \frac{2 t d t}{(t^{2} + 1) t} = 2 t a n^{- 1} t + c = 2 t a n^{- 1} (\sqrt{x + \frac{1}{x} + 1}) + c$
f(x) = $x + \frac{1}{x} + 1$

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Similar questions

Q. 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).

Q. 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).

Q. Find the HCF of

f (x) = x^{3} + x^{2} - x - 1

and

g (x) = x^{3} + x^{2} + x + 1

Q. (i) f(x) = x⁴

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62x² + 120x + 9

(ii) f(x) = x³

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6x² + 9x + 15

(iii) f(x) = (x

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1) (x+2)²

(iv) f(x) = 2/x

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2/x² , x>0

(v) f(x) = xe^x

(vi) f(x) = x/2+2/x, x>0

(vii) f(x) = (x+1) (x+2)^1/3,

x \frac{>}{} - 2

(viii) f(x) =

x \sqrt{32 - x^{2}}, - 5 \frac{<}{} x \frac{<}{} 5

(ix) f(x) =

x^{3} - 2 a x^{2} + a^{2} x, a > 0, x \in R

(x) f(x) =

x + \frac{a 2}{x}, a > 0,

x ≠ 0

(xi) f(x) =

x \sqrt{2 - x^{2}} - \sqrt{2} \leq x \leq \sqrt{2}

(xii) f(x) =

x + \sqrt{1 - x}, x \leq 1

Q. Find

f (3)

when

f (x) = x^{3} + x^{2} + x + 1

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

The correct option is A x+1x+1I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Put x+1+1x=t2(1−1x2)dx=2t dt=∫2t dt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c f(x) = x+1x+1

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).