Question

tentukan integral berikut ini​

Answer 1

Verified answer

Jawab:

[tex]\displaystyle \ln(x^2+4x+20)-\frac{\tan^{-1}\left ( \frac{x+2}{4} \right )}{4}+C[/tex]

Penjelasan dengan langkah-langkah:

Manipulasi aljabar pembilang 2x + 3 = 2x + 4 - 1 dan diselesaikan dengan teknik substitusi.

[tex]\displaystyle \int \frac{2x+3}{x^2+4x+20}~dx\\=\int \frac{2x+4-1}{x^2+4x+20}~dx\\=\int \left ( \frac{2x+4}{x^2+4x+20}-\frac{1}{x^2+4x+20} \right )dx\\=\int \frac{2x+4}{x^2+4x+20}~dx-\int \frac{dx}{x^2+4x+20}[/tex]

Integralkan [tex]\displaystyle \int \frac{2x+4}{x^2+4x+20}~dx[/tex]

[tex]\displaystyle \int \frac{2x+4}{x^2+4x+20}~dx\\u=x^2+4x+20\\du=2x+4~dx\\=\int \frac{2x+4}{u}~\frac{du}{2x+4}\\=\int \frac{1}{u}~du\\=\ln|u|\\=\ln(x^2+4x+20)[/tex]

Integralkan [tex]\displaystyle \int \frac{dx}{x^2+4x+20}~dx[/tex]

Cek diskriminan penyebut apakah bisa difaktorkan?

x² + 4x + 20

D = 4² - 4(1)(20) = -64

Akar-akar nya imajiner. Gunakan teknik kuadrat sempurna dan metode substitusi

[tex]\displaystyle \int \frac{dx}{x^2+4x+20}~dx\\=\int \frac{dx}{x^2+4x+4+16}~dx\\=\int \frac{dx}{(x+2)^2+4^2}\\=\int \frac{\frac{1}{4^2}}{\frac{(x+2)^2+4^2}{4^2}}~dx\\=\frac{1}{4^2}\int \frac{dx}{\frac{(x+2)^2}{4^2}+\frac{4^2}{4^2}}\\=\frac{1}{16}\int \frac{dx}{\left ( \frac{x+2}{4} \right )^2+1}[/tex]

[tex]\displaystyle v=\frac{x+2}{4}\\dv=\frac{1}{4}~dx[/tex]

[tex]\displaystyle =\frac{1}{16}\int \frac{4~dv}{v^2+1}\\=\frac{1}{4}\int \frac{dv}{v^2+1}\\=\frac{1}{4}\tan^{-1}v\\=\frac{1}{4}\tan^{-1}\left ( \frac{x+2}{4} \right )[/tex]

Jadi

[tex]\displaystyle \int \frac{2x+3}{x^2+4x+20}~dx\\=\int \frac{2x+4}{x^2+4x+20}~dx-\int \frac{dx}{x^2+4x+20}\\=\ln(x^2+4x+20)-\frac{1}{4}\tan^{-1}\left ( \frac{x+2}{4} \right )+C[/tex]