The value of -x2-1x4-x2-1dx is

The correct option is A tan^ 1(x^2 1x)+C Let I =∫x^2+1x^4 x^2+1dx=∫1+1/x^2x^2+1/x^2 1dx =∫(1+1/x^2)(x 1/x)^2+1dx Put x 1x = t ⇒(1+1x^2) dx= ...

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Standard XII

Mathematics

Integration by Substitution

The value of ...

Question

The value of $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$ is

{tan}^{- 1} (2 x^{2} - 1) + C

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{tan}^{- 1} (\frac{x^{2} - 1}{x}) + C

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{sin}^{- 1} (x - \frac{1}{x}) + C

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{tan}^{- 1} x^{2} + C

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Solution

The correct option is A ${tan}^{- 1} (\frac{x^{2} - 1}{x}) + C$
Let $I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 1} d x$
$= \int \frac{(1 + \frac{1}{x^{2}})}{{(x - \frac{1}{x})}^{2} + 1} d x$
Put $x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t$
$∴ I = \int \frac{d t}{t^{2} + 1} = {tan}^{- 1} t + C = {tan}^{- 1} (x - \frac{1}{x}) + C$
$= {tan}^{- 1} (\frac{x^{2} - 1}{x}) + C$

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Similar questions

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
If

0 < x < 1

, then

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

is equal to
(a)

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

(b)

x

(c)

x \sqrt{1 + x^{2}}

(d)

\sqrt{1 + x^{2}}

Q. The value of

\int_{- 1}^{3} [{t a n}^{- 1} (\frac{x}{x^{2} + 1}) + {t a n}^{- 1} (\frac{x^{2} + 1}{x})] d x =

\int x \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} d x =

Q. The value of integral

\int \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} [{({sec}^{- 1} \sqrt{1 + x^{2}})}^{2} + {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] d x

(x > 0),

is
(where

^{'} C^{'}

is the constant of integration)

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

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The value of ∫x2+1x4−x2+1dx is

The correct option is A tan−1(x2−1x)+CLet I=∫x2+1x4−x2+1dx=∫1+1x2x2+1x2−1dx=∫(1+1x2)(x−1x)2+1dxPut x−1x=t⇒(1+1x2)dx=dt∴I=∫dtt2+1=tan−1t+C=tan−1(x−1x)+C=tan−1(x2−1x)+C

The value of $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$ is