Pertanyaan

Selesaikanlah. ∫ t 4 ( t 2 + 1 ) 2 d t

Selesaikanlah.

$\int \frac{( t ^{2} + 1 ) ^{2}}{t ^{4}} d t$

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Jawaban

hasil dari adalah

hasil dari begin mathsize 14px style integral open parentheses t squared plus 1 close parentheses squared over t to the power of 4 space d t end style

begin mathsize 14px style integral open parentheses t squared plus 1 close parentheses squared over t to the power of 4 space d t end style

adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 over denominator 3 t cubed end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator 5 t to the power of 5 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table$

Pembahasan

Ingat bahwa: Aturan pangkat: Rumus integral: Untuk soal di atas kita dapat memanipulasi bentuk pembagiannya: Jadi, hasil dari adalah

Ingat bahwa:

Aturan pangkat:

a to the power of n times a to the power of m equals a to the power of n plus m end exponent 1 over a to the power of n equals a to the power of negative n end exponent

Rumus integral:

$integral a x to the power of n space d x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n minus 1 end exponent d x$

integral k space d x equals x comma space k equals k o n t a s n t a

integral open parentheses f open parentheses x close parentheses plus g open parentheses x close parentheses close parentheses space d x equals integral f open parentheses x close parentheses space d x plus g open parentheses x close parentheses space d x

Untuk soal di atas kita dapat memanipulasi bentuk pembagiannya:

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses t squared plus 1 close parentheses squared over t to the power of 4 space d t end cell equals cell integral fraction numerator t to the power of 4 plus 2 t squared plus 1 over denominator t to the power of 4 end fraction d t end cell row blank equals cell integral open parentheses open parentheses space t to the power of 4 plus 2 t squared plus 1 close parentheses t to the power of negative 4 end exponent close parentheses d t space space open parentheses i n g a t space 1 over t to the power of 4 equals t to the power of negative 4 end exponent close parentheses end cell row blank equals cell integral open parentheses space t to the power of 4 times t to the power of negative 4 end exponent plus 2 t squared times t to the power of negative 4 end exponent plus t to the power of negative 4 end exponent close parentheses space d t end cell row blank equals cell integral open parentheses t to the power of 4 minus 4 end exponent plus 2 t to the power of 2 minus 4 end exponent plus t to the power of negative 4 end exponent close parentheses d t end cell row blank equals cell integral open parentheses t to the power of 0 plus 2 t to the power of negative 2 end exponent plus t to the power of negative 4 end exponent close parentheses d t end cell row blank equals cell integral open parentheses 1 plus 2 t to the power of negative 2 end exponent plus t to the power of negative 4 end exponent close parentheses d t end cell row blank equals cell t plus fraction numerator 2 over denominator negative 3 end fraction t to the power of negative 3 end exponent plus fraction numerator 1 over denominator negative 5 end fraction t to the power of negative 5 end exponent plus c end cell row blank equals cell t minus fraction numerator 2 over denominator 3 t cubed end fraction minus fraction numerator 1 over denominator 5 t to the power of 5 end fraction plus c end cell end table$

Jadi, hasil dari begin mathsize 14px style integral open parentheses t squared plus 1 close parentheses squared over t to the power of 4 space d t end style adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 over denominator 3 t cubed end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator 5 t to the power of 5 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table$