Evaluate -x2-x2 - 1dx

The correct option is A x tan^ 1x+C =∫ x ^ 2 x ^ 2 +1 #160; dx =∫ x ^ 2 +1 1 x ^ 2 +1 #160; dx =∫dx ∫dxx^2+1 =x tan^ 1x+c ...

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Derivative of Standard Functions

Evaluate ∫x2...

Question

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

x - {cot}^{- 1} x + C

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

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

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None of these

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Solution

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\int \frac{e^{x} (x^{2} {tan}^{- 1} x + {tan}^{- 1} x + 1) d x}{x^{2} + 1}

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

\int_{0}^{1} \frac{{tan}^{- 1} x}{x^{2} + 1} d x =

{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

0 < x < 1

\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}

\frac{x}{\sqrt{1 + x^{2}}}

x

x \sqrt{1 + x^{2}}

\sqrt{1 + x^{2}}

{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

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

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