-01tan -11- x - x 2 dx -A- In 2B- -4C- -2D- -2- In 2

The correct option is A In 2∫_0^1tan^ 1 (1 x+x^2)dx = ∫_0^1 ( π/2 tan^ 1(x+(1 x)/1 x(1 x)) )dx = π/2 ∫^1_0 (tan^ 1 (1 x)+ tan^ 1 x)dx ⇒π/2 2 ∫_0^1 tan ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫01tan -11- x...

Question

$\int_{0}^{1} t a n^{- 1} (1 - x + x^{2}) d x =$ ___

I n 2

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\frac{π}{4}

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\frac{π}{2}

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\frac{π}{2} - I n 2

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Solution

The correct option is A $I n 2$
$\int_{0}^{1} t a n^{- 1} (1 - x + x^{2}) d x = \int_{0}^{1} (\frac{π}{2} - t a n^{- 1} (\frac{x + (1 - x)}{1 - x (1 - x)})) d x = \frac{π}{2} - \int_{0}^{1} (t a n^{- 1} (1 - x) + t a n^{- 1} x) d x \Rightarrow \frac{π}{2} - 2 \int_{0}^{1} t a n^{- 1} (x) d x = l n 2$

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∫10tan−1(1−x+x2)dx= ___

The correct option is A In 2∫10tan−1(1−x+x2)dx=∫10(π2−tan−1(x+(1−x)1−x(1−x)))dx=π2−∫10(tan−1(1−x)+tan−1x)dx⇒π2−2∫10tan−1(x)dx=ln 2

$\int_{0}^{1} t a n^{- 1} (1 - x + x^{2}) d x =$ ___