Evaluate- - x 4 -1 1- x 6 dx

The correct option is D tan ^ 1 (x)+ 1 / 3 tan ^ 1 ( x ^ 3 )+c Given, ∫x^4+1x^6+1dx =∫x^4+1x^6+1× #160; x^2+1x^2+1dx =∫(x^6+1)+x^2(x^2+1 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Evaluate: ∫...

Question

Evaluate: $\int \frac{x^{4} + 1}{1 + x^{6}} d x$

{t a n}^{- 1} (x) - {t a n}^{- 1} (x^{3}) + c

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{t a n}^{- 1} (x) - \frac{1}{3} {t a n}^{- 1} (x^{3}) + c

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{t a n}^{- 1} (x) + {t a n}^{- 1} (x^{3}) + c

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{t a n}^{- 1} (x) + \frac{1}{3} {t a n}^{- 1} (x^{3}) + c

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Solution

The correct option is D ${t a n}^{- 1} (x) + \frac{1}{3} {t a n}^{- 1} (x^{3}) + c$
Given,

$\int \frac{x^{4} + 1}{x^{6} + 1} d x$

$= \int \frac{x^{4} + 1}{x^{6} + 1} \times \frac{x^{2} + 1}{x^{2} + 1} d x$

$= \int \frac{(x^{6} + 1) + x^{2} (x^{2} + 1)}{(x^{6} + 1) (x^{2} + 1)} d x$

$= \int \frac{d x}{x^{2} + 1} + \frac{1}{3} \int \frac{3 x^{2}}{x^{6} + 1} d x$

$= {tan}^{- 1} x + \frac{1}{3} 3 \cdot \int \frac{x^{2}}{x^{6} + 1} d x$

substitute $u = x^{3}$

$= {tan}^{- 1} x + \frac{1}{3} 3 \cdot \int \frac{1}{3 (u^{2} + 1)} d u$

$= {tan}^{- 1} x + \frac{1}{3} 3 \cdot \frac{1}{3} \cdot \int \frac{1}{u^{2} + 1} d u$

$= {tan}^{- 1} x + \frac{1}{3} 3 \cdot \frac{1}{3} \cdot {tan}^{- 1} u$

$= {tan}^{- 1} x + \frac{1}{3} 3 \cdot \frac{1}{3} \cdot {tan}^{- 1} x^{3}$

$= {tan}^{- 1} x + \frac{1}{3} {tan}^{- 1} x^{3} + C$

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Similar questions

Q. The value of the integral

\int \frac{x^{4} + 1}{x^{6} + 1} d x

(C

is a constant of integration

)

Q. Assertion :

\int \frac{x^{4} + 1}{x^{6} + 1} d x = {tan}^{- 1} x + \frac{1}{3} {tan}^{- 1} x^{3} + C

Reason:

\int \frac{cos x + x sin x}{x (x + cos x)} d x = log (x^{2} + x cos x) + C

Q. Solve the following equations for x:
(i)

(ii)
(iii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

, x > 0
(vii)

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

(viii)
(ix)

Q. 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(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

, x > 0
(vii)

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

(viii)

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

(ix)

\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

(x)

\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}

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

.
(a)

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

(b)

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

0 \leq x \leq 1

then

(a, b) =

(a)

(0, π)

(b)

(0, π / 4)

(c)

(- π / 4, π / 4)

(d)

(π / 4, π / 2)

(d) If

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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