Roboguru

Tentukan integral tak tentu berikut: h.

Pertanyaan

Tentukan integral tak tentu berikut:

h. integral open parentheses x cubed plus 6 x to the power of 5 close parentheses plus open parentheses 3 x squared plus 6 close parentheses d x

Pembahasan Soal:

Rumus dasar integral yaitu:

  • integral k x to the power of n space d x equals fraction numerator k over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus straight C dengan sayarat straight n not equal to negative 1
  • integral k space d x equals k x plus straight C, suatu konstanta

Diperoleh penyelesaiannya yaitu:

integral open parentheses x cubed plus 6 x to the power of 5 close parentheses plus open parentheses 3 x squared plus 6 close parentheses d x equals integral x cubed space d x plus integral 6 x to the power of 5 space d x plus integral 3 x squared space d x plus integral 6 space d x equals fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent plus fraction numerator 6 over denominator 5 plus 1 end fraction x to the power of 5 plus 1 end exponent plus fraction numerator 3 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus 6 x plus straight C equals 1 fourth x to the power of 4 plus 6 over 6 x to the power of 6 plus 3 over 3 x cubed plus 6 x plus straight C equals 1 fourth x to the power of 4 plus x to the power of 6 plus x cubed plus 6 x plus straight C

Dengan demikian, integral tak tentu dari integral open parentheses x cubed plus 6 x to the power of 5 close parentheses plus open parentheses 3 x squared plus 6 close parentheses d x adalah 1 fourth x to the power of 4 plus x to the power of 6 plus x cubed plus 6 x plus straight C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Ayu

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

Hitunglah pengintegralan di bawah ini! 4)

Pembahasan Soal:

Integral fungsi begin mathsize 14px style f left parenthesis x right parenthesis equals a x to the power of n end style dapat ditentukan sebagai berikut.

begin mathsize 14px style integral space a x to the power of n space d x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end style  

sehingga integral dari fungsi yang diberikan di atas dapat ditentukan sebagai berikut.

begin mathsize 14px style integral space 2 x squared minus 3 x minus 2 space straight d x equals 2 over 3 x cubed minus 3 over 2 x cubed minus 2 x plus C end style  

0

Roboguru

Hasil dari  adalah .....

Pembahasan Soal:

Perhatikan perhitungan berikut!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 2 x squared plus square root of x plus 1 over x cubed close parentheses d x end cell equals cell integral 2 x squared blank d x plus integral square root of x blank d x plus integral 1 over x cubed blank d x end cell row blank equals cell integral 2 x squared blank d x plus integral x to the power of 1 half end exponent blank d x plus integral x to the power of negative 3 end exponent blank d x end cell row blank equals cell open parentheses fraction numerator 2 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C subscript 1 close parentheses plus open parentheses fraction numerator 1 over denominator 1 half plus 1 end fraction blank x to the power of 1 half plus 1 end exponent plus C subscript 2 close parentheses plus open parentheses fraction numerator 1 over denominator negative 3 plus 1 end fraction blank x to the power of negative 3 plus 1 end exponent plus C subscript 3 close parentheses end cell row blank equals cell 2 over 3 x cubed plus C subscript 1 plus fraction numerator 1 over denominator 3 over 2 end fraction x to the power of 3 over 2 end exponent plus C subscript 2 plus fraction numerator 1 over denominator negative 2 end fraction x to the power of negative 2 end exponent plus C subscript 3 end cell row blank equals cell 2 over 3 x cubed plus 2 over 3 x square root of x minus 1 half times 1 over x squared plus C subscript 1 plus C subscript 2 plus C subscript 3 end cell row blank equals cell 2 over 3 x cubed plus 2 over 3 x square root of x minus fraction numerator 1 over denominator 2 x squared end fraction plus C subscript 1 plus C subscript 2 plus C subscript 3 end cell end table end style 

Misalkan undefined maka

begin mathsize 14px style 2 over 3 x cubed plus 2 over 3 x square root of x minus fraction numerator 1 over denominator 2 x squared end fraction plus C subscript 1 plus C subscript 2 plus C subscript 3 equals 2 over 3 x cubed plus 2 over 3 x square root of x minus fraction numerator 1 over denominator 2 x squared end fraction plus C end style 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Tentukan integral-integral berikut ini. e.

Pembahasan Soal:

integral daribegin mathsize 14px style integral open parentheses x minus 2 close parentheses squared d x end style dengan menggunakan rumus-rumus dasar integral tak tentu.

Ingat,

integral a x to the power of n space straight d x equals fraction numerator a over denominator n plus 1 end fraction. x to the power of n plus 1 end exponent plus c space space space comma space n not equal to negative 1 

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses x minus 2 close parentheses squared d x end cell equals cell integral open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses d x end cell row blank equals cell integral x squared minus 4 x plus 4 d x end cell row blank equals cell integral x squared d x minus integral 4 x d x plus integral 4 d x end cell row blank equals cell 1 third x cubed minus 4 over 2 x squared plus 4 x plus C end cell row blank equals cell 1 third x cubed minus 2 x squared plus 4 x plus C end cell end table end style 

Dengan demikian integral dari undefined adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 third end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank cubed end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank squared end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank C end table.

0

Roboguru

Pembahasan Soal:

Integral space fungsi space polinom space adalah space  integral a x to the power of n d x space equals space fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C    Gunakan space formula space diatas space untuk space menyelesaikan space integral space dalam space soal space  integral open parentheses 1 half x cubed plus x plus 5 x squared close parentheses d x  equals open parentheses 1 half. fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent close parentheses plus open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent close parentheses plus open parentheses fraction numerator 5 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent close parentheses plus C  equals 1 over 8 x to the power of 4 plus 1 half x squared plus 5 over 3 x cubed plus C

0

Roboguru

Dengan menjabarkan bentuk aljabar dari integral, temukan hasil integrasi di bawah ini.

Pembahasan Soal:

Rumus integral adalah sebagai berikut:

axndx=n+1axn+1+c 

Penyelesaiannya adalah sebagai berikut:

(1x3)2dx=====(1x3)(1x3)dx(12x3+x6)dxx3+12x3+1+6+11x6+1+cx42x4+71x7+cx21x4+71x7+c 

Jadi, hasil integrasi begin mathsize 14px style integral open parentheses 1 minus x cubed close parentheses squared space d x end style adalah x21x4+71x7+c.

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