Bank Soal Matematika SMA Integral Substitusi

Soal

Pilgan

Hasil dari (4x51)2x63x dx\int\left(4x^5-1\right)\sqrt{2x^6-3x}\ dx adalah ....

A

29(2x63x)2x63x+C\frac{2}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

B

20x4(2x63x)2x63x+C20x^4\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

C

2x(2x63x)2x63x+C2x\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

D

19(2x63x)2x63x+C\frac{1}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

E

9(2x63x)2x63x+C9\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

Pembahasan:

Misalkan u=2x63xu=2x^6-3x, maka du=12x53dxdu=12x^5-3 dx dx=du12x53\Leftrightarrow dx=\frac{du}{12x^5-3}

Sehingga menjadi:

(4x51)2x63x dx=(2x63x)12 (4x51)dx\int\left(4x^5-1\right)\sqrt{2x^6-3x}\ dx=\int\left(2x^6-3x\right)^{\frac{1}{2}}\ \left(4x^5-1\right)dx

=u12 (4x51)du12x53=\int u^{\frac{1}{2}}\ \left(4x^5-1\right)\frac{du}{12x^5-3}

=u12 (4x51)du3(4x51)=\int u^{\frac{1}{2}}\ \left(4x^5-1\right)\frac{du}{3\left(4x^5-1\right)}

=u12 du3=\int u^{\frac{1}{2}}\ \frac{du}{3}

=13u12 du=\frac{1}{3}\int u^{\frac{1}{2}}\ du, 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

=13(112+1u12+1)+C=\frac{1}{3}\left(\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\right)+C

=13(112+22u12+22)+C=\frac{1}{3}\left(\frac{1}{\frac{1}{2}+\frac{2}{2}}u^{\frac{1}{2}+\frac{2}{2}}\right)+C

=13(132u32)+C=\frac{1}{3}\left(\frac{1}{\frac{3}{2}}u^{\frac{3}{2}}\right)+C

=13(23u32)+C=\frac{1}{3}\left(\frac{2}{3}u^{\frac{3}{2}}\right)+C

=29(u22)(u12)+C=\frac{2}{9}\left(u^{\frac{2}{2}}\right)\left(u^{\frac{1}{2}}\right)+C

=29uu+C=\frac{2}{9}u\sqrt{u}+C

=29(2x63x)2x63x+C=\frac{2}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C


Jadi, hasil integral substitusi tersebut adalah 29(2x63x)2x63x+C\frac{2}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C

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