If - x 2 tan - 1 x - 1 - x 2 d x - tan - 1 x - 1 - 2 log 1 - x 2 - f x - c then f x -

The correct option is C 1 / 2 ( tan ^ 1 x ) ^ 2 ∫>((1+x^2 1)tan^ 1x/1+x^2)dx =∫>(1 (1/1+x^2))tan^ 1x>dx =∫>1.tan^ 1dx ∫&g ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If ∫ x 2 t...

Question

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 $f (x) =$

- \frac{{tan}^{- 1} x}{2}

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

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\frac{{tan}^{- 1} x}{2}

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None of these

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Solution

The correct option is C $- \frac{1}{2} {({tan}^{- 1} x)}^{2}$
$\int (\frac{(1 + x^{2} - 1) t a n^{- 1} x}{1 + x^{2}}) d x = \int (1 - (\frac{1}{1 + x^{2}})) t a n^{- 1} x d x = \int 1. t a n^{- 1} d x - \int (\frac{t a n^{- 1} x}{1 + x^{2}}) d x U s e I L A T E i n f i r s t p a r t a n d a s s u m e t a n^{- 1} x = t f o r s e c o n d i n t e g r a t i o n = x t a n^{- 1} x - \int (\frac{1}{1 + x^{2}}) \times x d x - \int t d t = x t a n^{- 1} x - (\frac{1}{2}) l o g | 1 + x^{2} | - (\frac{t^{2}}{2}) + C = x t a n^{- 1} x - (\frac{1}{2}) l o g | 1 + x^{2} | - (\frac{(t a n^{- 1} x)^{2}}{2}) + C ∴ C = - (\frac{1}{2}) (t a n^{- 1} x)^{2}$

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Similar questions

Q. If

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

each is equal to

2 {tan}^{- 1} x .

,then show that

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

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