-13 Tan-1x-x2-1-Tan-1x2-1-x dx-

∫_ 1^3 ( Tan^ 1x/x^2+1+Tan^ 1x^2+1/x )dx=∫_ 1^0 [ Tan^ 1x/x^2+1+Tan^ 1x^2+1/x ]dx+∫_0^3 [ Tan^ 1x/x^2+1+Tan^ 1x^2+1/x ]dx=∫_ 1^0 π/2dx+∫^3_0 π/2dx= π/2+π/2(3 ( ...

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Byju's Answer

Standard XI

Mathematics

Improper Integrals

∫-13 Tan-1x/x...

Question

$\int_{- 1}^{3} (T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}) d x =$

π

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2 π

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4 π

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n o n e

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Solution

The correct option is A $π$
$\int_{- 1}^{3} (T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}) d x = \int_{- 1}^{0} [T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}] d x + \int_{0}^{3} [T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}] d x = \int_{- 1}^{0} - \frac{π}{2} d x + \int_{0}^{3} \frac{π}{2} d x = - \frac{π}{2} + \frac{π}{2} (3 - (0)) = π$

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Similar questions

\int_{- 1}^{3} (T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}) d x =

Q. Evaluate

\int_{- 1}^{3} ({tan}^{- 1} \frac{x}{x^{2} + 1} + {tan}^{- 1} \frac{x^{2} + 1}{x}) d x

Q. The value of

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

is equal to

Q. Assertion :

\int_{- 1}^{3} {{tan}^{- 1} (\frac{x}{1 + x^{2}}) + {tan}^{- 1} (\frac{x^{2} + 1}{x})} d x = 2 π

Reason:

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

and

{tan}^{- 1} t = {cot}^{- 1} \frac{1}{t}

\int_{- 1}^{3} (tan^{- 1} \frac{x}{x^{2} + 1} + tan^{- 1} \frac{x^{2} + 1}{x}) d x

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∫3−1(Tan−1xx2+1+Tan−1x2+1x)dx=

The correct option is A π∫3−1(Tan−1xx2+1+Tan−1x2+1x)dx=∫0−1[Tan−1xx2+1+Tan−1x2+1x]dx+∫30[Tan−1xx2+1+Tan−1x2+1x]dx=∫0−1−π2dx+∫30π2dx=−π2+π2(3−(0))=π

$\int_{- 1}^{3} (T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}) d x =$