Bank Soal Matematika SMA Integral Substitusi

Soal

Pilgan

x(x4)5dx=....\int x\left(x-4\right)^5dx=....

A

15(3x+2)(x4)6+C\frac{1}{5}\left(3x+2\right)\left(x-4\right)^6+C

B

16(3x+2)(x4)6+C\frac{1}{6}\left(3x+2\right)\left(x-4\right)^6+C

C

121(x4)6+C\frac{1}{21}\left(x-4\right)^6+C

D

121(3x+2)6+C\frac{1}{21}\left(3x+2\right)^6+C

E

121(3x+2)(x4)6+C\frac{1}{21}\left(3x+2\right)\left(x-4\right)^6+C

Pembahasan:

Misalkan u=x4u=x-4 atau x=u+4x=u+4, maka du=1dxdu=1 dx dx=du\Leftrightarrow dx={du}

Sehingga menjadi:

x(x4)5dx=(u+4)(u5)du\int x\left(x-4\right)^5dx=\int\left(u+4\right)\left(u^5\right)du

=(u6+4u5)du=\int\left(u^6+4u^5\right)du

=u6du+4u5du=\int u^6du+\int4u^5du, untuk f(x)=axn, n1f\left(x\right)=ax^n,\ n\ne-1 maka axndx=an+1xn+1+C\int ax^ndx=\frac{a}{n+1}x^{n+1}+C

=16+1u6+1+45+1u5+1+C=\frac{1}{6+1}u^{6+1}+\frac{4}{5+1}u^{5+1}+C

=17u7+46u6+C=\frac{1}{7}u^7+\frac{4}{6}u^6+C

=17u7+23u6+C=\frac{1}{7}u^7+\frac{2}{3}u^6+C

=321u7+1421u6+C=\frac{3}{21}u^7+\frac{14}{21}u^6+C

=121(3u7+14u6)+C=\frac{1}{21}\left(3u^7+14u^6\right)+C

=121(3u+14)u6+C=\frac{1}{21}\left(3u+14\right)u^6+C

=121(3(x4)+14)(x4)6+C=\frac{1}{21}\left(3\left(x-4\right)+14\right)\left(x-4\right)^6+C

=121(3x+2)(x4)6+C=\frac{1}{21}\left(3x+2\right)\left(x-4\right)^6+C


Jadi, hasil integral substitusi tersebut adalah 121(3x+2)(x4)6+C\frac{1}{21}\left(3x+2\right)\left(x-4\right)^6+C

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23 Februari 2021
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