Question

Integral metode substitusi HOTS level
[tex]\displaystyle \int \frac{1}{x^2\sqrt[4]{(x^4+1)^3}}~dx[/tex]

Answer 1

Verified answer

Penjelasan dengan langkah-langkah:

Integral subs Metod's

[tex]\displaystyle \int \frac{1}{x^2\sqrt[4]{(x^4+1)^3}}~dx[/tex]

• [tex]\displaystyle u=\frac{1}{x}[/tex]
• [tex] x^4=\frac{1}{u^4}[/tex]
• [tex]u'=-\frac{1}{x^2}[/tex]
• [tex]\displaystyle du=-\frac{1}{x^2}~dx\Rightarrow~dx=-x^2~du[/tex]

Maka

[tex]\begin{aligned}\displaystyle \int \frac{1}{x^2\sqrt[4]{(x^4+1)^3}}~dx&\Rightarrow \int \frac{1}{\cancel{x^2}\sqrt[4]{(\frac{1}{u^4}+1)^3}}~ \times -\cancel{x^2}~du\\& = \int - \frac{1}{\sqrt[4]{(\frac{1}{u^4}+1)^3}}~du \\& =\int \: - \frac{1}{\left( \frac{1 + {u}^{4} }{ {u}^{4} } \right) ^{ \frac{3}{4} } }~du \\& = \int - \frac{ {u}^{3} }{(1 + {u}^{4})^{ \frac{1}{4} } }~du \end{aligned}[/tex]

Misal

• v = u⁴
• v' = 4u³
• subtitusi du = [tex]\frac{1}{4u^3}~dv[/tex]

Maka

[tex]\begin{aligned}\int - \frac{ {u}^{3} }{(1 + {u}^{4})^{ \frac{1}{4} } }~du & \Rightarrow \int - \frac{ {u}^{3} }{(1 + v)^{ \frac{1}{4} } } \times \frac{1}{ {4u}^{3} }~dv \\ & = - \frac{1}{4} \int \frac{1}{(1 + v {)}^{ \frac{1}{4} } } ~dv \\& = - \frac{1}{4} \times \: 4 (1 + v {)}^{ \frac{1}{4} } \\ & = - \left(1 + \frac{1}{ {x}^{4} }\right) ^{ \frac{1}{4} }+C \end{aligned}[/tex]