Question

Tentukan hasil dari [tex]\int x - \sqrt{x + 1} \:dx [/tex]
[tex] \: [/tex]
​

Answer 1

Jawaban:

[tex]\int x - \sqrt{x + 1}= \rm \dfrac{1}{2} {x}^{2} - \dfrac{2}{3} {(x + 1)}^{ \frac{3}{2} } + C[/tex]

Penjelasan dengan langkah-langkah:

[tex]\int x - \sqrt{x + 1}[/tex]

[tex]x = \dfrac{1}{1 + 1} {x}^{1 + 1} = \dfrac{1}{2} {x}^{2} [/tex]

[tex] \rm \sqrt{x + 1} = {(x + 1)}^{ \frac{1}{2} } \rm = \frac{1}{ \frac{1}{2} + 1} {(x + 1)}^{ \frac{1}{2} + 1} = \dfrac{2}{3} {(x + 1)}^{ \frac{3}{2} } [/tex]

Maka:

[tex]\int x - \sqrt{x + 1}[/tex]

[tex] = \dfrac{1}{2} {x}^{2} - \dfrac{2}{3} {(x + 1)}^{ \frac{3}{2} } + C[/tex]

Answer 2

Verified answer

Jawab:

Penjelasan dengan langkah-langkah:

[tex]\int\ {x-\sqrt{x+1} } \, dx = \int\ {x-(x+1)^{\frac{1}{2} } \, dx[/tex]

For example :

u = x + 1

[tex]\frac{du}{dx} = 1\\du = dx[/tex]

[tex]\int\ {x-(x+1)^{\frac{1}{2} } \, dx = \int\ {x} dx - \int\ {(x+1)^{\frac{1}{2} } \, dx \\= \int\ {x} dx - \int\ {u^{\frac{1}{2} } \, du \\= \frac{1}{1+1} x^{1+1} - \frac{1}{\frac{1}{2} +1} u^{\frac{1}{2} +1}\\[/tex]

[tex]=\frac{1}{2} x^{2} - \frac{1}{\frac{3}{2} } u^{\frac{1}{2}+1}[/tex]

[tex]=\frac{1}{2}x^{2} - \frac{2}{3} u^{\frac{3}{2} }[/tex]

[tex]=\frac{1}{2}x^{2} - \frac{2}{3} (x+1)^{\frac{3}{2} }[/tex]

[tex]=\frac{1}{2}x^{2} - \frac{2}{3} (x+1).(x+1)^{\frac{1}{2} } \\=\frac{1}{2}x^{2} - \frac{2}{3} (x+1).\sqrt{x+1} +c[/tex]