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Berikut ini adalah pertanyaan dari andika7274236 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

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Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Integral Parsial

$\boxed{\int {u} \, dv=u\cdot v -\int {v} \, du}$

7. $\int {x(2x-1)^{-3} } \, dx=\int {u} \, dv$

u = x

du/dx = 1

du = dx

$dv=(2x-1)^{-3}\\\\v=\int{(2x-1)^{-3}} \, dx \\\\v=\frac{1}{-2} \cdot \frac{1}{2}(2x-1)^{-2} \\\\v=-\frac{1}{4} (2x-1)^{-2}$

$\int {x(2x-1)^{-3} } \, dx\\\\=u\cdot v-\int {v} \, du\\\\=x\cdot (-\frac{1}{4})(2x-1)^{-2} -\int { (-\frac{1}{4})(2x-1)^{-2} } \, dx \\ \\=-\frac{x}{4}(2x-1)^{-2}+\frac{1}{4} \int {(2x-1)^{-2} } \, dx\\\\=-\frac{x}{4}(2x-1)^{-2}+\frac{1}{4}\cdot \frac{1}{-1}\cdot \frac{1}{2}(2x-1)^{-1} +C\\ \\=-\frac{x}{4}(2x-1)^{-2}-\frac{1}{8}(2x-1)^{-1} +C\\\\\boxed{=-\frac{x}{4(2x-1)^{2}} -\frac{1}{8(2x-1)} +C}$

8. $\int {x^{4} \sqrt{1+x^{2} } } \, dx=\int {u} \, dv$

u = x⁴

du/dx = 4x³

du = 4x³ dx

$dv=\sqrt{1+x^{2} } \\\\dv=(1+x^{2} )^{\frac{1}{2} } \\\\v=\int {(1+x^{2} )^{\frac{1}{2} } } \, dx \\\\v=\frac{1}{\frac{3}{2} } (1+x^{2} )^{\frac{3}{2} }\\\\v=\frac{2}{3} (1+x^{2} )^{\frac{3}{2}}$

$\int {x^{4} \sqrt{1+x^{2} } } \, dx\\\\=x^{4}\cdot \frac{2}{3} (1+x^{2} )^{\frac{3}{2}}-\int {\frac{2}{3} (1+x^{2} )^{\frac{3}{2}} 4x^{3} } \, dx\\\\=\frac{2x^{4} }{3} (1+x^{2} )^{\frac{3}{2}}-\frac{8}{3} \int {x^{3} (1+x^{2} )^{\frac{3}{2}} } \, dx$

Ada integral parsial lagi di sini, maka dimisalkan jadi u dv lagi

$\int {x^{3} (1+x^{2} )^{\frac{3}{2}} } \, dx$

u = x³

du/dx = 3x²

du = 3x² dx

$dv= (1+x^{2} )^{\frac{3}{2}}\\\\v=\int { (1+x^{2} )^{\frac{3}{2}}} \, dx \\\\v=\frac{1}{\frac{5}{2} }\cdot \frac{1}{1} (1+x^{2} )^{\frac{5}{2}}\\\\v=\frac{2}{5} (1+x^{2} )^{\frac{5}{2}}$

$\int {x^{3} (1+x^{2} )^{\frac{3}{2}} } \, dx\\\\= x^{3} \cdot \frac{2}{5} (1+x^{2} )^{\frac{5}{2}} -\int {\frac{2}{5} (1+x^{2} )^{\frac{5}{2}}} \, 3x^{2} dx\\\\=\frac{2x^{3} }{5} (1+x^{2} )^{\frac{5}{2}}-\frac{6}{5} \int {x^{2} (1+x^{2} )^{\frac{5}{2}}} \, dx$

Ada integral parsial lagi, dimisalkan jadi u dv

$\int {x^{2} (1+x^{2} )^{\frac{5}{2}}} \, dx$

u = x²

du/dx = 2x

du = 2x dx

$dv=(1+x^{2} )^{\frac{5}{2} } \\\\v=\int {(1+x^{2} )^{\frac{5}{2} } } \, dx \\\\v=\frac{1}{\frac{7}{2} } (1+x^{2} )^{\frac{7}{2} }\\\\v=\frac{2}{7} (1+x^{2} )^{\frac{7}{2} }$

$\int {x^{2} (1+x^{2} )^{\frac{5}{2}}} \, dx\\\\=x^{2} \cdot \frac{2}{7} (1+x^{2} )^{\frac{7}{2} } -\int {\frac{2}{7} (1+x^{2} )^{\frac{7}{2} }} 2x\, dx \\\\=\frac{2x^{2} }{7} (1+x^{2} )^{\frac{7}{2} }-\frac{4}{7} \int {x(1+x^{2} )^{\frac{7}{2} }} \, dx$

Sekali lg, terdapat integral parsial lagi:)

$\int {x(1+x^{2} )^{\frac{7}{2} }} \, dx$

u = x

du/dx = 1

du = dx

$dv=(1+x^{2} )^{\frac{7}{2} }\\\\v=\int {(1+x^{2} )^{\frac{7}{2} }} \, dx \\\\v=\frac{2}{9}(1+x^{2} )^{\frac{9}{2} }$

$\int {x(1+x^{2} )^{\frac{7}{2} }} \, dx\\\\=x\cdot \frac{2}{9}(1+x^{2} )^{\frac{9}{2} }-\int {\frac{2}{9}(1+x^{2} )^{\frac{9}{2} }} \, dx \\\\=\frac{2x}{9} (1+x^{2} )^{\frac{9}{2} }-\frac{2}{9} \int {(1+x^{2} )^{\frac{9}{2} }} \, dx\\\\=\frac{2x}{9} (1+x^{2} )^{\frac{9}{2} }-\frac{2}{9}\cdot\frac{2}{11} (1+x^{2} )^{\frac{11}{2} }+C$

$=\frac{2x}{9} (1+x^{2} )^{\frac{9}{2} }-\frac{4}{99}(1+x^{2} )^{\frac{11}{2} }+C$

Akhirnya dpt 1 hasilnya, substitusi ke integral parsial sblm ini

$\int {x^{2} (1+x^{2} )^{\frac{5}{2}}} \, dx\\\\=\frac{2x^{2} }{7} (1+x^{2} )^{\frac{7}{2} }-\frac{4}{7} \int {x(1+x^{2} )^{\frac{7}{2} }} \, dx\\\\=\frac{2x^{2} }{7} (1+x^{2} )^{\frac{7}{2} }-\frac{4}{7}\cdot \frac{2x}{9} (1+x^{2} )^{\frac{9}{2} }-\frac{4}{7}\cdot \frac{4}{99}(1+x^{2} )^{\frac{11}{2} }+C\\\\=\frac{2x^{2} }{7} (1+x^{2} )^{\frac{7}{2} }-\frac{8x}{63} (1+x^{2} )^{\frac{9}{2} }-\frac{16}{693 }(1+x^{2} )^{\frac{11}{2} }+C$

Kita substitusi lagi ke integral parsial sblmnya

$\int {x^{3} (1+x^{2} )^{\frac{3}{2}} } \, dx\\\\=\frac{2x^{3} }{5} (1+x^{2} )^{\frac{5}{2}}-\frac{6}{5} \int {x^{2} (1+x^{2} )^{\frac{5}{2}}} \, dx\\\\=\frac{2x^{3} }{5} (1+x^{2} )^{\frac{5}{2}}-\frac{6}{5} \cdot \frac{2x^{2} }{7} (1+x^{2} )^{\frac{7}{2} }-\frac{6}{5}\cdot \frac{8x}{63} (1+x^{2} )^{\frac{9}{2} }-\frac{6}{5}\cdot \frac{16}{693 }(1+x^{2} )^{\frac{11}{2} }+C$

$=\frac{2x^{3} }{5} (1+x^{2} )^{\frac{5}{2}}-\frac{12x^{2} }{35} (1+x^{2} )^{\frac{7}{2} }-\frac{16x}{105} (1+x^{2} )^{\frac{9}{2} }-\frac{32}{1155} (1+x^{2} )^{\frac{11}{2} }+C$

Substitusi lagii ke sblmnya, plg awal

$\int {x^{4} \sqrt{1+x^{2} } } \, dx\\\\=\frac{2x^{4} }{3} (1+x^{2} )^{\frac{3}{2}}-\frac{8}{3} \int {x^{3} (1+x^{2} )^{\frac{3}{2}} } \, dx\\\\= \frac{2x^{4} }{3} (1+x^{2} )^{\frac{3}{2}}-\frac{8}{3} (\frac{2x^{3} }{5} (1+x^{2} )^{\frac{5}{2}}-\frac{12x^{2} }{35} (1+x^{2} )^{\frac{7}{2} }-\frac{16x}{105} (1+x^{2} )^{\frac{9}{2} }-\frac{32}{1155} (1+x^{2} )^{\frac{11}{2} })+C$

$\boxed{= \frac{2x^{4} }{3} (1+x^{2} )^{\frac{3}{2}}- \frac{16x^{3} }{15} (1+x^{2} )^{\frac{5}{2}}-\frac{32x^{2} }{35} (1+x^{2} )^{\frac{7}{2} }-\frac{128x}{315} (1+x^{2} )^{\frac{9}{2} }-\frac{256}{3465} (1+x^{2} )^{\frac{11}{2} }+C}$