The correct option is B #10; x +log|x^2+1|+4tan^ 1x+c #10; #160;∫ (x^2+2x+5/x^2+1)dx #160;∫ [1+2(x+2)/(x^2+1)]dx #160;∫ 1· dx+2∫x/x^2+ ...

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Byju's Answer

Standard XII

Mathematics

Functions

∫x2+2x+5/x2+1...

Question

$\int \frac{x^{2} + 2 x + 5}{x^{2} + 1} d x =$

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

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x - log (x^{2} + 1) - {tan}^{- 1} x + c

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x + log | x^{2} + 1 | + 4 {tan}^{- 1} x + c

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

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Solution

The correct option is B $x + log | x^{2} + 1 | + 4 {tan}^{- 1} x + c$
$\int (\frac{x^{2} + 2 x + 5}{x^{2} + 1}) d x$
$\int [1 + \frac{2 (x + 2)}{(x^{2} + 1)}] d x$
$\int 1 \cdot d x + 2 \int \frac{x}{x^{2} + 1} d x + \int \frac{4}{x^{2} + 1} d x$
$x + log (x^{2} + 1) + 4 {tan}^{- 1} x + c$

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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. If

I = \int \frac{x^{2} - 2}{(x^{4} + 5 x^{2} + 4) t a n^{- 1} (\frac{x^{2} + 2}{x})} d x = \frac{A}{368} log | t a n^{- 1} (x + 2 / x) | + C

then A equals.

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

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∫x2+2x+5x2+1dx=

The correct option is B x+log|x2+1|+4tan−1x+c ∫(x2+2x+5x2+1)dx∫[1+2(x+2)(x2+1)]dx∫1⋅dx+2∫xx2+1dx+∫4x2+1dxx+log(x2+1)+4tan−1x+c

$\int \frac{x^{2} + 2 x + 5}{x^{2} + 1} d x =$

The correct option is B $x + log | x^{2} + 1 | + 4 {tan}^{- 1} x + c$
$\int (\frac{x^{2} + 2 x + 5}{x^{2} + 1}) d x$
$\int [1 + \frac{2 (x + 2)}{(x^{2} + 1)}] d x$
$\int 1 \cdot d x + 2 \int \frac{x}{x^{2} + 1} d x + \int \frac{4}{x^{2} + 1} d x$
$x + log (x^{2} + 1) + 4 {tan}^{- 1} x + c$