- x 4- x 4 dx is equal toa 14 tan -1 x 2- Cb 14 tan -1 - 22c 12 tan -1 - 22d none of these

(b) 14tan 1 #160;x22 Let #160;I= #8747;x4+x4dx #160; #160; #160; #160; #160; #160; #160; #160;= #8747;x #160;dx22+x22Putting #160;x ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

∫ x 4+ x 4 dx...

Question

$\int \frac{x}{4 + x^{4}} d x$ is equal to

(a) $\frac{1}{4} \tan^{- 1} x^{2} + C$

(b) $\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})$

(c) $\frac{1}{2} \tan^{- 1} (\frac{x^{2}}{2})$

(d) none of these

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Solution

(b) $\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})$

$Let I = \int \frac{x}{4 + x^{4}} d x = \int \frac{x d x}{2^{2} + {(x^{2})}^{2}} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{2^{2} + t^{2}} = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{t}{2}) + C (∵ \int \frac{1}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a}) = \frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2}) + C (∵ t = x^{2})$

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Similar questions

If $x_{1}, x_{2}, x_{3}, x_{4}$ are roots of the equation $x^{4} - x^{3}$ $s i n 2 β + x^{2} c o s 2 β - x c o s β - s i n β = 0 t h e n \sum_{i = 1}^{4} t a n^{- 1} x_{i}$ is equal to

\int (x + x^{2} + x^{3} + x^{4})

dx is equal to

\int \frac{1}{7 + 5 \cos x} d x =

(a)

\frac{1}{\sqrt{6}} \tan^{- 1} (\frac{1}{\sqrt{6}} \tan \frac{x}{2}) + C

(b)

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan \frac{x}{2}) + C

(c)

\frac{1}{4} \tan^{- 1} (\tan \frac{x}{2}) + C

(d)

\frac{1}{7} \tan^{- 1} (\tan \frac{x}{2}) + C

\frac{1}{\sqrt{9} - \sqrt{8}}

is equal to

(a)

3 + 2 \sqrt{2}

(b)

\frac{1}{3 + 2 \sqrt{2}}

(c)

3 - 2 \sqrt{2}

(d)

\frac{3}{2} - \sqrt{2}

Q. If

x_{1}, x_{2}, x_{3}, x_{4},

are roots of the equation

x^{4} - x^{3} sin 2 β + x^{2} cos 2 β - x cos β - sin β = 0

, then

4 \sum i = 1 {tan}^{- 1} x_{i}

is equal to

(

where

β \in (0, \frac{π}{2}) - {\frac{π}{6}})

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∫x4+x4 dx is equal to (a) 14tan-1 x2+C (b) 14tan-1 x22 (c) 12tan-1 x22 (d) none of these

(b) 14tan-1 x22 Let I=∫x4+x4dx =∫x dx22+x22Putting x2=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dt22+t2 =12×12 tan-1 t2+C ∵∫1a2+x2=1atan-1xa =14 tan-1 x22+C ∵t =x2

$\int \frac{x}{4 + x^{4}} d x$ is equal to

(a) $\frac{1}{4} \tan^{- 1} x^{2} + C$

(b) $\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})$

(c) $\frac{1}{2} \tan^{- 1} (\frac{x^{2}}{2})$

(d) none of these