∫x2(x3+3)5dx= ....\int x^2\left(x^3+3\right)^5dx=\ ....∫x2(x3+3)5dx= ....
118(x3+3)5+C\frac{1}{18}\left(x^3+3\right)^5+C181(x3+3)5+C
118(x3+3)6+C\frac{1}{18}\left(x^3+3\right)^6+C181(x3+3)6+C
2x(x3+3)6+C2x\left(x^3+3\right)^6+C2x(x3+3)6+C
x2(x3+3)6+Cx^2\left(x^3+3\right)^6+Cx2(x3+3)6+C
12(x3+3)6+C\frac{1}{2}\left(x^3+3\right)^6+C21(x3+3)6+C
Misalkan u=x3+3u=x^3+3u=x3+3, maka du=3x2dxdu=3x^2dxdu=3x2dx ⇔dx=du3x2\Leftrightarrow dx=\frac{du}{3x^2}⇔dx=3x2du
Sehingga menjadi:
∫x2(x3+3)5dx=∫(x3+3)5x2dx\int x^2\left(x^3+3\right)^5dx=\int\left(x^3+3\right)^5x^2dx∫x2(x3+3)5dx=∫(x3+3)5x2dx
=∫u5x2du3x2=\int u^5x^2\frac{du}{3x^2}=∫u5x23x2du
=∫u5du3=\int u^5\frac{du}{3}=∫u53du
=13∫u5du=\frac{1}{3}\int u^5du=31∫u5du
=13(15+1u5+1)+C=\frac{1}{3}\left(\frac{1}{5+1}u^{5+1}\right)+C=31(5+11u5+1)+C
=13(16u6)+C=\frac{1}{3}\left(\frac{1}{6}u^6\right)+C=31(61u6)+C
=118u6+C=\frac{1}{18}u^6+C=181u6+C
=118(x3+3)6+C=\frac{1}{18}\left(x^3+3\right)^6+C=181(x3+3)6+C
Jadi, hasil integral substitusi tersebut adalah 118(x3+3)6+C\frac{1}{18}\left(x^3+3\right)^6+C181(x3+3)6+C