Question

Coba kerjakan ini dengan metode substitusi
[tex]\displaystyle \int (x^6+x^3)\sqrt[3]{x^3+2}~dx[/tex]

Answer 1

Verified answer

Penjelasan dengan langkah-langkah:

Integral subs Metod's

Cara 1

[tex]\displaystyle \int (x^6+x^3)\sqrt[3]{x^3+2}~dx[/tex]

• u = x³ + 1
• x = [tex]\displaystyle (u-1)^{\frac{1}{3}}[/tex]
• du = 3x² dx
• dx = [tex]\displaystyle\frac{1}{3x^2}~du[/tex]

Jabarkan

[tex]\begin{aligned}\displaystyle\int( {x}^{6} + {x}^{3}) \sqrt[3]{ {x}^{3} + 2} \: \: dx&=\int {x}^{3}( {x}^{3} + 1)( {x}^{3} + 1 + 1 {)}^{ \frac{1}{3} } \: \frac{1}{3 {x}^{2} } \: du \\&= \frac{1}{3} \int \: x(u)(u + 1 {)}^{ \frac{1}{3} } \: du \\&= \frac{1}{3} \int u(x - 1 {)}^{ \frac{1}{3} }(u + 1 {)}^{ \frac{1}{3} } \: du \\ &= \frac{1}{3} \int u( {u}^{2} - 1 {)}^{ \frac{1}{3} } \: du \end{aligned}[/tex]

misal

• v = u² - 1 = (x³+1)² - 1
• dv = 2u  du
• du = [tex]\displaystyle\frac{1}{2u}[/tex] dv

Maka

[tex]\begin{aligned}\displaystyle\frac{1}{3} \int u( {u}^{2} - 1 {)}^{ \frac{1}{3} } \: du &= \frac{1}{3}\int u(v {)}^{ \frac{1}{3} } \frac{1}{2u} \: dv \\ &= \frac{1}{6} \int \: (v {)}^{ \frac{1}{3} } \: dv \\ &= \frac{ {v}^{ \frac{4}{3} } }{8} \\&= \frac{ (( {x}^{3} + 1)^{2}-1 )^{ \frac{4}{3} }}{8} \\ &=\boxed{ \frac{ {x}^{4}( {x}^{3} + 2 {)}^{ \frac{4}{3} } }{8} + C} \end{aligned}[/tex]

Cara 2

Jabarkan

[tex]\begin{aligned}\displaystyle\int( {x}^{6} + {x}^{3}) \sqrt[3]{ {x}^{3} + 2} \: \: dx&=\int \sqrt[3]{x^3}(x^5+x^2) \sqrt[3]{ {x}^{3} + 2} ~dx \\&=\int\sqrt[3]{x^3}\sqrt[3]{ {x}^{3} + 2}~(x^5+x^2) ~dx \\&=\int\sqrt[3]{x^3}\sqrt[3]{ {x}^{3} + 2}~~x^2(x^3+1) ~dx \\&=\int\sqrt[3]{x^3+1-1}~\sqrt[3]{ {x}^{3} + 1 +1}~~x^2(x^3+1) ~dx \end{aligned}[/tex]

• u = x³ + 1

• du = 3x² dx
• dx = [tex]\displaystyle\frac{1}{3x^2} ~du[/tex]

Maka

[tex]\begin{aligned}\displaystyle\int\sqrt[3]{x^3+1-1}~\sqrt[3]{ {x}^{3} + 1 +1}~~x^2(x^3+1) ~dx&=\int \sqrt[3]{u-1}~ \sqrt[3]{u+1} ~x^2(u)~\frac{1}{3x^2} ~du \\&=\frac{1}{3} \int~u\sqrt[3]{u^2-1} ~du \end{aligned}[/tex]

• v = u² - 1
• dv = 2u du
• du = [tex]\displaystyle\frac{1}{2u}~ dv[/tex]

[tex]\begin{aligned}\displaystyle\frac{1}{3} \int~u\sqrt[3]{u^2-1} ~du&=\frac{1}{3}\int~u\sqrt[3]{v}~\frac{1}{2u}~dv\\&=\frac{1}{6}\int v^{\frac{1}{3}}~dv\\&=\frac{1}{6}\times\frac{v^{\frac{4}{3}}}{\frac{4}{3}}\\&\cdots\:ingat~v=u^2-1 , dan~u=x^3+1\\&=\frac{((x^3+1)^2-1)^{\frac{4}{3}}}{8} \\&=\boxed{ \frac{ {x}^{4}( {x}^{3} + 2 {)}^{ \frac{4}{3} } }{8} + C} \end{aligned}[/tex]