The questions is in the picture.help me, they're sucks

Berikut ini adalah pertanyaan dari Christopher1997 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

The questions is in the picture.
help me, they're sucks

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

part (b)

$\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx$

change this form to partial fraction

$\frac{12x^2+2x+3}{x(1+4x^2)} } =\frac{3+12x^2}{x(1+4x^2)} +\frac{2x}{x(1+4x^2)} =\frac{3}{x} +\frac{2}{1+4x^2}$

therefore

$\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=\int {\left(\frac{3}{x} +\frac{2}{1+4x^2} \right)} \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx$

suppose

$2x=\tan t \ \Rightarrow \ 2\ dx=\sec^2t\ dt$

then

$\int {\frac{2}{1+4x^2}} \, dx=\int {\frac{1}{1+\tan^2t}} \, \sec^2t\,dt=\int {\frac{sec^2t\,dt}{\sec^2t}}=\int {dt}=t+C$

since $\tan t=2x \ \Rightarrow \ t=\arctan(2x)$

$\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C$

hence

$\int {\frac{2}{1+4x^2}} \, dx=\frac{1}{2} \left(\frac{2x}{1+4x^2} +\arctan(2x) \right)+C$

$\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx=3\ln|x|+ \arctan(2x) +C$

done

part (c)

in progress...

from part (b), we get indefinite integral

$\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C$

$\int {\frac{1}{1+4x^2}} \, dx=\frac{1}{2} \arctan(2x)+C$

$= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b) \right)$

use the limit, so that the improper integral becomes a definite integral.

$\int\limits^\infty_0 {\frac{1}{1+4x^2} } \, dx = \lim_{b \to \infty} \left(\int\limits^b_0 {\frac{1}{1+4x^2} } \, dx \right)$

$= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b)-\arctan(2\cdot0)) \right)$

$= \lim_{b \to \infty} \left(\frac{1}{2} \arctan(2b) \right)$

$= \frac{1}{2}\lim_{b \to \infty} \left( \arctan(2b) \right)$

$=\frac{1}{2} \left(\frac{\pi }{2} \right)$

$=\frac{\pi}{4}$

because tha range of $y=\arctan x$ is

$R_y=\{y:\frac{-\pi }{2}$

attached proof

$part (b)[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx[/tex]change this form to partial fraction[tex]\frac{12x^2+2x+3}{x(1+4x^2)} } =\frac{3+12x^2}{x(1+4x^2)} +\frac{2x}{x(1+4x^2)} =\frac{3}{x} +\frac{2}{1+4x^2}[/tex]therefore[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=\int {\left(\frac{3}{x} +\frac{2}{1+4x^2} \right)} \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx[/tex]suppose[tex]2x=\tan t \ \Rightarrow \ 2\ dx=\sec^2t\ dt[/tex]then[tex]\int {\frac{2}{1+4x^2}} \, dx=\int {\frac{1}{1+\tan^2t}} \, \sec^2t\,dt=\int {\frac{sec^2t\,dt}{\sec^2t}}=\int {dt}=t+C[/tex]since [tex]\tan t=2x \ \Rightarrow \ t=\arctan(2x)[/tex][tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]hence[tex]\int {\frac{2}{1+4x^2}} \, dx=\frac{1}{2} \left(\frac{2x}{1+4x^2} +\arctan(2x) \right)+C[/tex][tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx=3\ln|x|+ \arctan(2x) +C[/tex]donepart (c)in progress...from part (b), we get indefinite integral[tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]so[tex]\int {\frac{1}{1+4x^2}} \, dx=\frac{1}{2} \arctan(2x)+C[/tex][tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b) \right)[/tex]use the limit, so that the improper integral becomes a definite integral.[tex]\int\limits^\infty_0 {\frac{1}{1+4x^2} } \, dx = \lim_{b \to \infty} \left(\int\limits^b_0 {\frac{1}{1+4x^2} } \, dx \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b)-\arctan(2\cdot0)) \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2} \arctan(2b) \right)[/tex] [tex]= \frac{1}{2}\lim_{b \to \infty} \left( \arctan(2b) \right)[/tex] [tex]=\frac{1}{2} \left(\frac{\pi }{2} \right)[/tex] [tex]=\frac{\pi}{4}[/tex]because tha range of [tex]y=\arctan x[/tex] is [tex]R_y=\{y:\frac{-\pi }{2} <y<\frac{\pi}{2} \}[/tex]attached proof$ $part (b)[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx[/tex]change this form to partial fraction[tex]\frac{12x^2+2x+3}{x(1+4x^2)} } =\frac{3+12x^2}{x(1+4x^2)} +\frac{2x}{x(1+4x^2)} =\frac{3}{x} +\frac{2}{1+4x^2}[/tex]therefore[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=\int {\left(\frac{3}{x} +\frac{2}{1+4x^2} \right)} \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx[/tex]suppose[tex]2x=\tan t \ \Rightarrow \ 2\ dx=\sec^2t\ dt[/tex]then[tex]\int {\frac{2}{1+4x^2}} \, dx=\int {\frac{1}{1+\tan^2t}} \, \sec^2t\,dt=\int {\frac{sec^2t\,dt}{\sec^2t}}=\int {dt}=t+C[/tex]since [tex]\tan t=2x \ \Rightarrow \ t=\arctan(2x)[/tex][tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]hence[tex]\int {\frac{2}{1+4x^2}} \, dx=\frac{1}{2} \left(\frac{2x}{1+4x^2} +\arctan(2x) \right)+C[/tex][tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx=3\ln|x|+ \arctan(2x) +C[/tex]donepart (c)in progress...from part (b), we get indefinite integral[tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]so[tex]\int {\frac{1}{1+4x^2}} \, dx=\frac{1}{2} \arctan(2x)+C[/tex][tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b) \right)[/tex]use the limit, so that the improper integral becomes a definite integral.[tex]\int\limits^\infty_0 {\frac{1}{1+4x^2} } \, dx = \lim_{b \to \infty} \left(\int\limits^b_0 {\frac{1}{1+4x^2} } \, dx \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b)-\arctan(2\cdot0)) \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2} \arctan(2b) \right)[/tex] [tex]= \frac{1}{2}\lim_{b \to \infty} \left( \arctan(2b) \right)[/tex] [tex]=\frac{1}{2} \left(\frac{\pi }{2} \right)[/tex] [tex]=\frac{\pi}{4}[/tex]because tha range of [tex]y=\arctan x[/tex] is [tex]R_y=\{y:\frac{-\pi }{2} <y<\frac{\pi}{2} \}[/tex]attached proof$ $part (b)[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx[/tex]change this form to partial fraction[tex]\frac{12x^2+2x+3}{x(1+4x^2)} } =\frac{3+12x^2}{x(1+4x^2)} +\frac{2x}{x(1+4x^2)} =\frac{3}{x} +\frac{2}{1+4x^2}[/tex]therefore[tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=\int {\left(\frac{3}{x} +\frac{2}{1+4x^2} \right)} \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx[/tex]suppose[tex]2x=\tan t \ \Rightarrow \ 2\ dx=\sec^2t\ dt[/tex]then[tex]\int {\frac{2}{1+4x^2}} \, dx=\int {\frac{1}{1+\tan^2t}} \, \sec^2t\,dt=\int {\frac{sec^2t\,dt}{\sec^2t}}=\int {dt}=t+C[/tex]since [tex]\tan t=2x \ \Rightarrow \ t=\arctan(2x)[/tex][tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]hence[tex]\int {\frac{2}{1+4x^2}} \, dx=\frac{1}{2} \left(\frac{2x}{1+4x^2} +\arctan(2x) \right)+C[/tex][tex]\int {\frac{12x^2+2x+3}{x(1+4x^2)} } \, dx=3\ln|x|+\int {\frac{2}{1+4x^2}} \, dx=3\ln|x|+ \arctan(2x) +C[/tex]donepart (c)in progress...from part (b), we get indefinite integral[tex]\int {\frac{2}{1+4x^2}} \, dx=\arctan(2x)+C[/tex]so[tex]\int {\frac{1}{1+4x^2}} \, dx=\frac{1}{2} \arctan(2x)+C[/tex][tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b) \right)[/tex]use the limit, so that the improper integral becomes a definite integral.[tex]\int\limits^\infty_0 {\frac{1}{1+4x^2} } \, dx = \lim_{b \to \infty} \left(\int\limits^b_0 {\frac{1}{1+4x^2} } \, dx \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2}( \arctan(2b)-\arctan(2\cdot0)) \right)[/tex] [tex]= \lim_{b \to \infty} \left(\frac{1}{2} \arctan(2b) \right)[/tex] [tex]= \frac{1}{2}\lim_{b \to \infty} \left( \arctan(2b) \right)[/tex] [tex]=\frac{1}{2} \left(\frac{\pi }{2} \right)[/tex] [tex]=\frac{\pi}{4}[/tex]because tha range of [tex]y=\arctan x[/tex] is [tex]R_y=\{y:\frac{-\pi }{2} <y<\frac{\pi}{2} \}[/tex]attached proof$

Semoga dengan pertanyaan yang sudah terjawab oleh atkoamatarp dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

Apabila terdapat kesalahan dalam mengerjakan soal, silahkan koreksi jawaban dengan mengirimkan email ke yomemimo.com melalui halaman Contact

Last Update: Fri, 01 Apr 22

<< Previous Post

Next Post >>

Kerjakan soal, tugas, & PR sekolah semakin mudah

The questions is in the picture.help me, they're sucks​

Jawaban dan Penjelasan

part (b)

part (c)

attached proof

Video Tutorial Android dan Smartphone

Pertanyaan Lain Mata pelajaran Matematika

The questions is in the picture.help me, they're sucks