Pertanyaan

Diketahui . Jika F(x) = ∫ f(x) dx dan , maka F(x) = ...

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Misalkan dan Didapat bahwa dan Karena F(x) = ∫ f(x) dx, maka didapat Kemudian selesaikan bentuk dengan integral substitusi trigonometri. Misalkan Didapat bahwa dan Jadi, Maka didapat Kemudian diketahui pula bahwa , maka Maka diperoleh, Jadi, jawaban yang benar adalah B.

Misalkan

dan

$begin mathsize 14px style fraction numerator d v over denominator d x end fraction equals 1 end style$

Didapat bahwa

$begin mathsize 14px style fraction numerator d u over denominator d x end fraction equals fraction numerator 1 over denominator 1 plus x squared end fraction times 2 x fraction numerator d u over denominator d x end fraction equals fraction numerator 2 x over denominator 1 plus x squared end fraction end style$

dan

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell integral fraction numerator d v over denominator d x end fraction d x end cell row blank equals cell integral 1 d x end cell row blank equals cell x plus C end cell end table end style$

Karena F(x) = ∫ f(x) dx, maka didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight F open parentheses x close parentheses end cell equals cell integral ln invisible function application open parentheses 1 plus x squared close parentheses d x end cell row blank equals cell integral u times fraction numerator d v over denominator d x end fraction d x end cell row blank equals cell u times v minus integral v times fraction numerator d u over denominator d x end fraction d x end cell row blank equals cell ln invisible function application open parentheses 1 plus x squared close parentheses times x minus integral x times fraction numerator 2 x over denominator open parentheses 1 plus x squared close parentheses end fraction blank d x end cell row blank equals cell x ln invisible function application open parentheses 1 plus x squared close parentheses minus integral fraction numerator 2 x squared over denominator open parentheses 1 plus x squared close parentheses end fraction blank d x end cell end table end style$

Kemudian selesaikan bentuk $begin mathsize 14px style integral fraction numerator 2 x squared over denominator open parentheses 1 plus x squared close parentheses end fraction blank d x end style$ dengan integral substitusi trigonometri. Misalkan

Didapat bahwa

dan

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d x over denominator d theta end fraction end cell equals cell sec squared invisible function application theta end cell row cell integral fraction numerator d x over denominator d theta end fraction d theta end cell equals cell integral sec squared invisible function application theta d theta end cell row cell integral d x end cell equals cell integral sec squared invisible function application theta d theta end cell end table end style$

Jadi,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 2 x squared over denominator open parentheses 1 plus x squared close parentheses end fraction blank d x end cell equals cell integral fraction numerator 2 tan squared invisible function application theta over denominator open parentheses 1 plus tan squared invisible function application theta close parentheses end fraction times sec squared invisible function application theta blank d theta end cell row blank equals cell integral fraction numerator 2 tan squared invisible function application theta over denominator sec squared invisible function application theta end fraction times sec squared invisible function application theta blank d theta end cell row blank equals cell integral 2 tan squared invisible function application theta blank d theta end cell row blank equals cell 2 integral tan squared invisible function application theta d theta end cell row blank equals cell 2 integral open parentheses sec squared invisible function application theta minus 1 close parentheses d theta end cell row blank equals cell 2 open parentheses tan invisible function application theta minus straight theta close parentheses plus C end cell row blank equals cell 2 tan invisible function application theta minus 2 theta plus C end cell row blank equals cell 2 x minus 2 tan to the power of negative 1 end exponent invisible function application x plus C end cell end table end style$

Maka didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight F open parentheses x close parentheses end cell equals cell x ln invisible function application open parentheses 1 plus x squared close parentheses minus integral fraction numerator 2 x squared over denominator open parentheses 1 plus x squared close parentheses end fraction blank d x end cell row blank equals cell x ln invisible function application open parentheses 1 plus x squared close parentheses minus open parentheses 2 x minus 2 tan to the power of negative 1 end exponent invisible function application x close parentheses plus C end cell row blank equals cell x ln invisible function application open parentheses 1 plus x squared close parentheses minus 2 x plus 2 tan to the power of negative 1 end exponent invisible function application x plus C end cell end table end style$

Kemudian diketahui pula bahwa , maka

Maka diperoleh,

Jadi, jawaban yang benar adalah B.