Question

Q. (1/11)

Tentukan integral dari ..........
[tex]\displaystyle\int~\frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}~dx[/tex]
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Answer 1

Verified answer

Jawab:

[tex]\displaystyle \arcsin\left ( \frac{x}{\sqrt[4]{1+x^4}} \right )+C[/tex]

Penjelasan dengan langkah-langkah:

Substitusi [tex]\displaystyle x=\sqrt{\tan\theta}\rightarrow dx=\frac{\sec^2\theta}{2\sqrt{\tan\theta}}~d\theta[/tex]

[tex]\begin{aligned}&\int \frac{dx}{(1+x^4)\sqrt{\sqrt{1+x^4}}-x^2}\\&=\int \frac{\frac{\sec^2\theta}{2\sqrt{\tan\theta}}}{(1+\tan^2\theta)\sqrt{\sqrt{1+\tan^2\theta}}-\tan\theta}~d\theta\\&=\int \frac{\frac{\sec^2\theta}{2\sqrt{\tan\theta}}}{\sec^2\theta\sqrt{\sqrt{\sec^2\theta}}-\tan\theta}~d\theta\\&=\frac{1}{2}\int \frac{d\theta}{\sqrt{\tan\theta}\sqrt{\sec\theta-\tan\theta}}\end{aligned}[/tex]

[tex]\begin{aligned}&=\frac{1}{2}\int \frac{d\theta}{\sqrt{\frac{\sin\theta}{\cos\theta}}\sqrt{\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}}\\&=\frac{1}{2}\int \frac{d\theta}{\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}~\frac{\sqrt{1-\sin\theta}}{\sqrt{\cos\theta}}}\\&=\frac{1}{2}\int \frac{\cos\theta}{\sqrt{\sin\theta}\sqrt{1-\sin\theta}}~d\theta\end{aligned}[/tex]

Substitusi [tex]\displaystyle u=\sin\theta\rightarrow du=\cos\theta~d\theta[/tex]

[tex]\begin{aligned}&=\frac{1}{2}\int \frac{\cos\theta}{\sqrt{u}\sqrt{1-u}}~\frac{du}{\cos\theta}\\&=\frac{1}{2}\int \frac{du}{\sqrt{u}\sqrt{1-u}}\end{aligned}[/tex]

Substitusi [tex]\displaystyle v=\sqrt{u}\rightarrow dv=\frac{1}{2\sqrt{u}}~du[/tex]

[tex]\begin{aligned}&=\frac{1}{2}\int \frac{1}{v\sqrt{1-v^2}}~\frac{dv}{\frac{1}{2\sqrt{u}}}\\&=\int \frac{\sqrt{u}}{v\sqrt{1-v^2}}~dv\\&=\int \frac{v}{v\sqrt{1-v^2}}~dv\\&=\sin^{-1}v+C\\&=\sin^{-1}\sqrt{u}+C\\&=\sin^{-1}\sqrt{\sin\theta}+C\\&=\sin^{-1}\sqrt{\frac{x^2}{\sqrt{1+(x^2)^2}}}+C\\&=\sin^{-1}\left ( \frac{x}{\sqrt[4]{1+x^4}} \right )+C\end{aligned}[/tex]