Evaluate -8 x2 exx2tan -1 x-x2 -1 xwhere C is constant of integration

Let I=∫8 x^2 e^xx^2tan ^ 1 x+x^2 ^ 1 x ⇒ I=∫8 x^2· e^xx^2(tan ^ 1 x+ ^ 1 x) d x ⇒ I=∫8 e^x(π / 2) d x=16π∫ e^x d x (tan^ 1x+^ 1=π2;x∈ℝ) ⇒ I=16π e^x+C nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

Evaluate ∫8 x...

Question

Evaluate $\int \frac{8 x^{2} e^{x}}{x^{2} {tan}^{- 1} x + x^{2} {cot}^{- 1} x}$
(where $C$ is constant of integration)

\frac{16}{π} e^{x} + C

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$\frac{16}{π} e^{- x} + C$

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- \frac{16}{π} e^{x} + C

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- \frac{16}{π} e^{- x} + C

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Solution

The correct option is A $\frac{16}{π} e^{x} + C$
Let $I = \int \frac{8 x^{2} e^{x}}{x^{2} {tan}^{- 1} x + x^{2} {cot}^{- 1} x}$
$\Rightarrow I = \int \frac{8 x^{2} \cdot e^{x}}{x^{2} ({tan}^{- 1} x + {cot}^{- 1} x)} d x$
$\Rightarrow I = \int \frac{8 e^{x}}{(π / 2)} d x = \frac{16}{π} \int e^{x} d x$
$({tan}^{- 1} x + {cot}^{- 1} = \frac{π}{2}; x \in R)$

$\Rightarrow I = \frac{16}{π} e^{x} + C$

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Evaluate ∫8x2exx2tan−1x+x2cot−1x (where C is constant of integration)

The correct option is A 16πex+CLet I=∫8x2exx2tan−1x+x2cot−1x ⇒I=∫8x2⋅exx2(tan−1x+cot−1x)dx ⇒I=∫8ex(π/2)dx=16π∫exdx (tan−1x+cot−1=π2;x∈R) ⇒I=16πex+C

Evaluate $\int \frac{8 x^{2} e^{x}}{x^{2} {tan}^{- 1} x + x^{2} {cot}^{- 1} x}$
(where $C$ is constant of integration)