Answer:
[tex]\boxed{{ \:\displaystyle\int\bf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } }=- \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \: }} \\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } } \\ \\ [/tex]
To evaluate this integral, we substitute
[tex]\sf \: 1 + x = \dfrac{1}{y} \\ \\ [/tex]
[tex]\sf\implies dx = - \dfrac{1}{ {y}^{2} } \: dy \\ \\ [/tex]
Also,
[tex]\sf \: 1 + x = \dfrac{1}{y} \: \: \sf\implies x = \dfrac{1 - y}{y} \\ \\ [/tex]
So, on substituting these values in above integral, we get
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ - \dfrac{1}{ {y}^{2} } }{\dfrac{1}{y} \sqrt{3 + \dfrac{2(1 - y)}{y} - {\bigg(\dfrac{1 - y}{y} \bigg) }^{2} } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{y \sqrt{3 + \dfrac{2(1 - y)}{y} - \dfrac{ {(1 - y)}^{2} }{ {y}^{2} } } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{y \sqrt{\dfrac{ 3 {y}^{2} + 2y(1 - y) - {(1 - y)}^{2} }{ {y}^{2} } } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ 3 {y}^{2} + 2y(1 - y) - {(1 - y)}^{2} } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ 3 {y}^{2} + 2y - {2y}^{2} - (1 + {y}^{2} - 2y)} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ {y}^{2} + 2y - (1 + {y}^{2} - 2y)} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ {y}^{2} + 2y - 1 - {y}^{2} + 2y} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{4y - 1} } \: dy \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: \displaystyle\int\sf \frac{dx}{ \sqrt{x} } = 2 \sqrt{x} + c \: }} \\ \\ [/tex]
[tex]\sf \: = \: - \dfrac{2 \sqrt{4y - 1} }{4} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{4y - 1} }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{4}{x + 1} - 1} }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{4 - x - 1}{x + 1} } }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{3 - x }{x + 1} } }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \\ \\ [/tex]
Hence,
[tex]\boxed{ \sf{ \:\displaystyle\int\sf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } } \: = \: - \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \: }} \\ \\ [/tex]
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Verified answer
Answer:
[tex]\boxed{{ \:\displaystyle\int\bf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } }=- \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \: }} \\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } } \\ \\ [/tex]
To evaluate this integral, we substitute
[tex]\sf \: 1 + x = \dfrac{1}{y} \\ \\ [/tex]
[tex]\sf\implies dx = - \dfrac{1}{ {y}^{2} } \: dy \\ \\ [/tex]
Also,
[tex]\sf \: 1 + x = \dfrac{1}{y} \: \: \sf\implies x = \dfrac{1 - y}{y} \\ \\ [/tex]
So, on substituting these values in above integral, we get
[tex]\sf \: = \: \displaystyle\int\sf \dfrac{ - \dfrac{1}{ {y}^{2} } }{\dfrac{1}{y} \sqrt{3 + \dfrac{2(1 - y)}{y} - {\bigg(\dfrac{1 - y}{y} \bigg) }^{2} } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{y \sqrt{3 + \dfrac{2(1 - y)}{y} - \dfrac{ {(1 - y)}^{2} }{ {y}^{2} } } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{y \sqrt{\dfrac{ 3 {y}^{2} + 2y(1 - y) - {(1 - y)}^{2} }{ {y}^{2} } } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ 3 {y}^{2} + 2y(1 - y) - {(1 - y)}^{2} } } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ 3 {y}^{2} + 2y - {2y}^{2} - (1 + {y}^{2} - 2y)} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ {y}^{2} + 2y - (1 + {y}^{2} - 2y)} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{ {y}^{2} + 2y - 1 - {y}^{2} + 2y} } \: dy \\ \\ [/tex]
[tex]\sf \: = \: - \displaystyle\int\sf \dfrac{ 1 }{\sqrt{4y - 1} } \: dy \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: \displaystyle\int\sf \frac{dx}{ \sqrt{x} } = 2 \sqrt{x} + c \: }} \\ \\ [/tex]
[tex]\sf \: = \: - \dfrac{2 \sqrt{4y - 1} }{4} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{4y - 1} }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{4}{x + 1} - 1} }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{4 - x - 1}{x + 1} } }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\sqrt{\dfrac{3 - x }{x + 1} } }{2} + c \\ \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \\ \\ [/tex]
Hence,
[tex]\boxed{ \sf{ \:\displaystyle\int\sf \dfrac{dx}{(1 + x) \sqrt{3 + 2x - {x}^{2} } } \: = \: - \: \dfrac{1}{2} \sqrt{\dfrac{3 - x }{x + 1}} + c \: }} \\ \\ [/tex]