Question

Q.

[tex] \sf\int(8x - 12) \sqrt{ {x}^{2} - 3x - 1 \: } dx = ...[/tex]
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Answer 1

Jawaban:INTEGRAL SUBSTITUSIBentuk umum integral sebagai berikut: [tex] \int \: {ax}^{n} \rightarrow \: \frac{a {x}^{n + 1} }{n + 1} {}^{} + C \\ [/tex]Pembahasan[tex] \int(8x - 12) \sqrt{ {x}^{2} - 3x - 1 \: } dx [/tex] Misalkan: [tex]u = {x}^{2} - 3x - 1 \: \rightarrow \: {u}^{ ' } = 2x - 3[/tex][tex]dx = \frac{du}{2x - 3} \\ [/tex]Maka: [tex] \int \: (8x - 12) \: . \: {u}^{ \frac{1}{2} } \: . \: dx \\ [/tex][tex] \int \: (8x - 12). \: \frac{ {u}^{ \frac{1}{2} + 1} }{( \frac{1}{2} + 1)} . \: \frac{du}{(2x - 3)} \\ [/tex][tex] \int \: \frac{(8x - 12)}{(2x - 3)} . \: \frac{ {u}^{ \frac{3}{2} } }{( \frac{3}{2}) } \: . \: du \\ [/tex][tex] \int \: \frac{4 \cancel{(2x - 3)} }{ \cancel{(2x - 3)} } . \frac{1}{( \frac{3}{2}) } . {u}^{ \frac{3}{2} } \: . \: du \\ [/tex][tex] = 4. \: \frac{2}{3} . {u}^{ \frac{3}{2} } . \: du \\ [/tex][tex] = \frac{8}{3} . \: {u}^{ \frac{3}{2} } . \: du \\ [/tex][tex] = \frac{8}{3} .( {x}^{2} - 3x - 1) {}^{ \frac{3}{2} } + C \\ [/tex][tex] = \frac{8}{ 3} ( {x}^{2} - 3x - 1) \sqrt{ {x}^{2} - 3x - 1} + C \\ [/tex]Jadi, hasil dari [tex] \int(8x - 12) \sqrt{ {x}^{2} - 3x - 1 \: } dx [/tex] adalah [tex] \frac{8}{ 3} ( {x}^{2} - 3x - 1) \sqrt{ {x}^{2} - 3x - 1} + C [/tex].