Pertanyaan

Hasil dari ...

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Misalkan dan Oleh karena itu, didapat bahwa dan Kemudian perhatikan bahwa Maka jika kita buat sebuah segitiga siku-siku maka diperoleh Sehingga Maka integral pada soal menjadi Lalu selesaikan bentuk dengan integral subsitusi. Misalkan Akibat Jadi, jawaban yang benar adalah C.

Misalkan

dan

$begin mathsize 14px style fraction numerator d v over denominator d x end fraction equals e to the power of x end style$

Oleh karena itu, didapat bahwa

dan

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

Kemudian perhatikan bahwa

Maka jika kita buat sebuah segitiga siku-siku maka diperoleh

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell e to the power of x end cell equals cell tan invisible function application u end cell row cell fraction numerator d over denominator d x end fraction e to the power of x end cell equals cell fraction numerator d over denominator d x end fraction tan invisible function application u end cell row cell e to the power of x end cell equals cell sec squared invisible function application u times fraction numerator d u over denominator d x end fraction end cell row cell e to the power of x end cell equals cell fraction numerator 1 over denominator cos squared invisible function application u end fraction times fraction numerator d u over denominator d x end fraction end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell e to the power of x cos squared invisible function application u end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell e to the power of x open parentheses fraction numerator 1 over denominator square root of e to the power of 2 x end exponent plus 1 end root end fraction close parentheses squared end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell fraction numerator e to the power of x over denominator e to the power of 2 x end exponent plus 1 end fraction end cell end table end style$

Maka integral pada soal menjadi

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

Lalu selesaikan bentuk $begin mathsize 14px style integral fraction numerator e to the power of 2 x end exponent over denominator e to the power of 2 x end exponent plus 1 end fraction d x end style$ dengan integral subsitusi.

Misalkan

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row p equals cell e to the power of 2 x end exponent plus 1 end cell row cell fraction numerator d p over denominator d x end fraction end cell equals cell 2 e to the power of 2 x end exponent end cell end table end style$

Akibat

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

Jadi, jawaban yang benar adalah C.