Integrate the following- - - 1 x2 - x - 1dx

The correct option is B 1/2log( 1 + x^2/2 + x^2 + 2x) + tan ^ 1( x + 1) + C [ I = ∫_^^1( x^2 + x + 1)dx; ...

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Standard XII

Mathematics

Algebra of Derivatives

Integrate the...

Question

Integrate the following:
$\int {cot}^{- 1} (x^{2} + x + 1) d x$

x {cot}^{1} (x^{2} + x + 1) + C

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

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

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x {cot}^{- 1} (x^{2} + x + 1) + \frac{1}{2} log (\frac{1 + x^{2}}{2 + x^{2} + 2 x})

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Solution

The correct option is B $\frac{1}{2} log (\frac{1 + x^{2}}{2 + x^{2} + 2 x}) + {tan}^{- 1} (x + 1) + C$
$\begin{matrix} I = \int {cot}^{1} (x^{2} + x + 1) d x = \int {tan}^{- 1} (\frac{1}{x^{2} + x + 1}) d x = \int {tan}^{- 1} (\frac{1}{1 + x (1 + x)}) d x = \int ({tan}^{- 1} (x + 1) - {tan}^{- 1} ({tan}^{- 1} x)) d x = \int {tan}^{- 1} (x + 1) d x + \int {tan}^{- 1} (x) d x I_{1} = \int {tan}^{- 1} (x + 1) d x \end{matrix}$
$\begin{matrix} P u t x + 1 = t d x = d t I_{1} = \int {tan}^{- 1} t d t = t {tan}^{- 1} t - \int \frac{t}{1 + t^{2}} d t = t {tan}^{- 1} t - \frac{1}{2} log (1 + t^{2}) + C = (1 + x) {tan}^{- 1} (x + 1) - \frac{1}{2} log [1 + (1 + x^{2})] + C \end{matrix}$
$\begin{matrix} I_{2} = \int {tan}^{- 1} (x) d x = x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2}) + C \end{matrix}$

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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. If

\int \frac{x^{2} {tan}^{- 1} x}{1 + x^{2}} d x = {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2}) + f (x) + c

then

f (x) =

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

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)

equals

A. x tan⁻¹ (x + 1) + C

B. tan^−
1 (x + 1) + C

C. (x + 1) tan⁻¹ x + C

D. tan⁻¹ x + C

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Integrate the following:∫cot−1(x2+x+1)dx

Integrate the following:
$\int {cot}^{- 1} (x^{2} + x + 1) d x$