Answer:
[tex]\sf - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c \\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf (6x + 5) \sqrt{6 - {2x}^{2} + x} \: dx \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \displaystyle\int\sf (6x + 5) \sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left(\dfrac{6}{ - 4} ( - 4x + 1 - 1) + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1 - 1) + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{3}{2} + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{3 + 10}{2}\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{13}{2}\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: - \dfrac{3}{2} \displaystyle\int\sf ( - 4x + 1)\sqrt{ - {2x}^{2} + x + 6} dx + \dfrac{13}{2}\displaystyle\int\sf \sqrt{ { - 2x}^{2} + x + 6 }dx \\ \\ [/tex]
Let assume that
[tex]\sf \: = \: I_1 + I_2 \\ \\ [/tex]
where,
[tex]\sf \: I_1= \: - \dfrac{3}{2} \displaystyle\int\sf ( - 4x + 1)\sqrt{ - {2x}^{2} + x + 6} dx \\ \\ [/tex]
and
[tex]\sf \: I_2 = \dfrac{13}{2}\displaystyle\int\sf \sqrt{ { - 2x}^{2} + x + 6 }dx \\ \\ [/tex]
Now, Consider
Substitute,
[tex]\sf \: \sqrt{ - {2x}^{2} + x + 6} = y \\ \\ [/tex]
[tex]\sf \: - {2x}^{2} + x + 6 = {y}^{2} \\ \\ [/tex]
[tex]\sf\implies \sf \: ( - 4x + 1)dx \: = \: 2y \: dy \\ \\ [/tex]
So, using these values, above integral can be rewritten as
[tex]\sf \: I_1 = - \dfrac{3}{2}\displaystyle\int\sf y \times 2y \: dy \\ \\ [/tex]
[tex]\sf \: I_1 = -3\displaystyle\int\sf {y}^{2} \: dy \\ \\ [/tex]
[tex]\sf \: I_1 = -3 \times \dfrac{ {y}^{3} }{3} + c_1 \\ \\ [/tex]
[tex]\sf \: I_1 = - {y}^{3} + c_1 \\ \\ [/tex]
[tex]\sf\implies \sf \: I_1 = - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + c_1 \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13}{2}\displaystyle\int\sf \sqrt{ - 2\left( {x}^{2} - \dfrac{x}{2} - 3 \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ - \left( {x}^{2} - \dfrac{x}{2} + \dfrac{1}{16} - \dfrac{1}{16} - 3 \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ - \left( {\bigg( x - \dfrac{1}{4} \bigg) }^{2} - \dfrac{49}{16} \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} }dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{\bigg( x - \dfrac{1}{4} \bigg)}{2} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2\\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2\\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \displaystyle\int\sf (6x + 5) \sqrt{6 - {2x}^{2} + x} \: dx \\ \\ [/tex]
[tex]\sf \: = \: - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + c_1 + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2 \\ \\ [/tex]
[tex]\sf = - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c \\ \\ [/tex]
[tex]\sf \: c = c_1 + c_2 \\ \\ [/tex]
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Verified answer
Answer:
[tex]\sf - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c \\ \\ [/tex]
Step-by-step explanation:
Given integral is
[tex]\sf \: \displaystyle\int\sf (6x + 5) \sqrt{6 - {2x}^{2} + x} \: dx \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \displaystyle\int\sf (6x + 5) \sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left(\dfrac{6}{ - 4} ( - 4x + 1 - 1) + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1 - 1) + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{3}{2} + 5\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{3 + 10}{2}\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: \displaystyle\int\sf \left( - \dfrac{3}{2} ( - 4x + 1) + \dfrac{13}{2}\right)\sqrt{ - {2x}^{2} + x + 6} \: dx \\ \\ [/tex]
[tex]\sf \: = \: - \dfrac{3}{2} \displaystyle\int\sf ( - 4x + 1)\sqrt{ - {2x}^{2} + x + 6} dx + \dfrac{13}{2}\displaystyle\int\sf \sqrt{ { - 2x}^{2} + x + 6 }dx \\ \\ [/tex]
Let assume that
[tex]\sf \: = \: I_1 + I_2 \\ \\ [/tex]
where,
[tex]\sf \: I_1= \: - \dfrac{3}{2} \displaystyle\int\sf ( - 4x + 1)\sqrt{ - {2x}^{2} + x + 6} dx \\ \\ [/tex]
and
[tex]\sf \: I_2 = \dfrac{13}{2}\displaystyle\int\sf \sqrt{ { - 2x}^{2} + x + 6 }dx \\ \\ [/tex]
Now, Consider
[tex]\sf \: I_1= \: - \dfrac{3}{2} \displaystyle\int\sf ( - 4x + 1)\sqrt{ - {2x}^{2} + x + 6} dx \\ \\ [/tex]
Substitute,
[tex]\sf \: \sqrt{ - {2x}^{2} + x + 6} = y \\ \\ [/tex]
[tex]\sf \: - {2x}^{2} + x + 6 = {y}^{2} \\ \\ [/tex]
[tex]\sf\implies \sf \: ( - 4x + 1)dx \: = \: 2y \: dy \\ \\ [/tex]
So, using these values, above integral can be rewritten as
[tex]\sf \: I_1 = - \dfrac{3}{2}\displaystyle\int\sf y \times 2y \: dy \\ \\ [/tex]
[tex]\sf \: I_1 = -3\displaystyle\int\sf {y}^{2} \: dy \\ \\ [/tex]
[tex]\sf \: I_1 = -3 \times \dfrac{ {y}^{3} }{3} + c_1 \\ \\ [/tex]
[tex]\sf \: I_1 = - {y}^{3} + c_1 \\ \\ [/tex]
[tex]\sf\implies \sf \: I_1 = - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + c_1 \\ \\ [/tex]
Now, Consider
[tex]\sf \: I_2 = \dfrac{13}{2}\displaystyle\int\sf \sqrt{ { - 2x}^{2} + x + 6 }dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13}{2}\displaystyle\int\sf \sqrt{ - 2\left( {x}^{2} - \dfrac{x}{2} - 3 \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ - \left( {x}^{2} - \dfrac{x}{2} + \dfrac{1}{16} - \dfrac{1}{16} - 3 \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ - \left( {\bigg( x - \dfrac{1}{4} \bigg) }^{2} - \dfrac{49}{16} \right)}dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2}\displaystyle\int\sf \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} }dx \\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{\bigg( x - \dfrac{1}{4} \bigg)}{2} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2\\ \\ [/tex]
[tex]\sf \: I_2 = \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2\\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \displaystyle\int\sf (6x + 5) \sqrt{6 - {2x}^{2} + x} \: dx \\ \\ [/tex]
[tex]\sf \: = \: - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + c_1 + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c_2 \\ \\ [/tex]
[tex]\sf = - { (\sqrt{ - {2x}^{2} + x + 6 } ) \: }^{3} + \dfrac{13 \sqrt{2} }{2} \left[ \dfrac{4x - 1}{8} \sqrt{ {\bigg( \dfrac{7}{4}\bigg) }^{2} - {\bigg( x - \dfrac{1}{4} \bigg) }^{2} } + \dfrac{49}{32} {sin}^{ - 1} \dfrac{4x - 1}{7} \right] + c \\ \\ [/tex]
where,
[tex]\sf \: c = c_1 + c_2 \\ \\ [/tex]