soal tersebut soal integral substitusi tolong di jawab

<p>Jawaban &nbsp;&nbsp;<strong>- 1/(32 (4x² + 16x + 5)⁸) + C</strong></p><p>&nbsp;</p><p><strong>Pembahasan&nbsp;</strong></p><p>∫ (2x + 4)/(4x² + 16x + 5)⁹dx</p><p>&nbsp;</p><ul><li>Metode substitusi :</li></ul><p>&nbsp; &nbsp; &nbsp; &nbsp;u = 4x² + 16x + 5</p><p>maka&nbsp;</p><p>&nbsp; &nbsp; du = (8x + 16) dx</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 4(2x + 4) dx</p><p>&lt;=&gt; (2x + 4) dx = ¼ du</p><p>&nbsp;</p><p>Ubah persamaan integral tersebut dengan substitusi menjadi :</p><p>&nbsp;∫ &nbsp;¼ du. 1/u⁠⁠⁠⁠⁠⁠⁠⁹= ¼ ∫ u^-9du</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= ¼. -⅛ u^-8⁠⁠ + C</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= - 1/(32 u⁸) + C</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = - 1/(32 (4x² + 16x + 5)⁸ + C</p><p>&nbsp;</p><p>&nbsp; &nbsp; &nbsp;&nbsp;</p><p>&nbsp;</p>