Question

$\large \displaystyle \sf\int_{0}^{ \infty } \frac{ {r}^{n - 1} }{( {r}^{2} + 1) ^{ \frac{n + 2}{2} } }$​

Answer 1

$\large{\boxed{\mathtt{\fcolorbox{black}{pink}{꧁Question꧂}}}}$

$\large \displaystyle \sf\int_{0}^{ \infty } \frac{ {r}^{n - 1} }{( {r}^{2} + 1) ^{ \frac{n + 2}{2} } }$

$\large{\underline{\mathtt{\red{♡︎A}\pink{N}\green{S}\blue{W}\red{E}\purple{R♡︎}}}}$

Refer to the attachment!

Answer 2

Answer:

$\huge \green{ \frac{1}{n} }$

Step-by-step explanation:

$\displaystyle \sf\int_{0}^{ \infty } \frac{ {r}^{n - 1} }{( {r}^{2} + 1) ^{ \frac{n + 2}{2} } }dr \\ \\ = \displaystyle \sf\int_{0}^{ \infty } \frac{ {r}^{n - 1} }{( {r}^{2} ) ^{ \frac{n + 2}{2} } (1+ \frac{1}{ { r}^{2} } ) ^{ \frac{n + 2}{2} } }dr \\ \\ = \displaystyle \sf\int_{0}^{ \infty } \frac{ {r}^{n - 1} }{ {r}^{n + 2} ( 1+ \frac{1}{ {r}^{2} } ) ^{ \frac{n + 2}{2} } } dr\\ \\ = \displaystyle \sf\int_{0}^{ \infty } \frac{ 1 }{ {r}^{3} ( 1 + \frac{1}{ {r}^{2} } ) ^{ \frac{n + 2}{2} } }dr \\ \\ put \: \: 1 + \frac{1}{ {r}^{2} } = x \implies \: \frac{ - 2dr}{ {r}^{3} } = dx \\ \\ = \displaystyle \sf\int_{ \infty}^{ 1} \frac{ - dx}{2( {x})^{ \frac{n + 2}{2} } } \\ \\ = \frac{1}{2} \bigg[ \frac{x {}^{ - \frac{n + 2}{2} + 1 } }{ - \frac{n + 2}{2} + 1 } \bigg] _{ \infty }^{ 1} \\ \\ = - \frac{1}{2} \bigg[ \frac{1 - 0} { \frac{ - n - 2 + 2}{2} }\bigg] \\ \\ = - \frac{1}{2} \times \frac{2}{ - n} = \frac{1}{n}$