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begin mathsize 14px style integral fraction numerator left parenthesis x plus 1 right parenthesis left parenthesis x minus 3 right parenthesis over denominator square root of x end fraction d x equals end style....

  1. begin mathsize 14px style 2 over 5 x squared square root of x minus 4 over 3 x square root of x minus 6 square root of x plus C end style   

  2. size 14px 2 over size 14px 5 size 14px x to the power of size 14px 2 square root of size 14px x size 14px minus size 14px 4 over size 14px 3 size 14px x square root of size 14px x size 14px plus size 14px 6 square root of size 14px x size 14px plus size 14px C      

  3. size 14px 2 over size 14px 5 size 14px x to the power of size 14px 2 square root of size 14px x size 14px plus size 14px 4 over size 14px 3 size 14px x square root of size 14px x size 14px minus size 14px 6 square root of size 14px x size 14px plus size 14px C        

  4. size 14px 2 over size 14px 5 size 14px x to the power of size 14px 2 square root of size 14px x size 14px plus size 14px 4 over size 14px 3 size 14px x square root of size 14px x size 14px plus size 14px 6 square root of size 14px x size 14px plus size 14px C     

  5. size 14px 1 over size 14px 2 size 14px x to the power of size 14px 2 square root of size 14px x size 14px minus size 14px 2 over size 14px 3 size 14px x square root of size 14px x size 14px minus size 14px 3 square root of size 14px x size 14px plus size 14px C    

Pembahasan Video:

Pembahasan Soal:

Integral tak tentu merupakan invers/kebalikan dari turunan. Karena integral dan turunan berkaitan, maka rumus integral dapat diperoleh dari rumusan penurunan. Jika turunan:

fraction numerator straight d over denominator straight d x end fraction fraction numerator a over denominator left parenthesis n plus 1 right parenthesis end fraction x to the power of left parenthesis n plus 1 right parenthesis end exponent equals a times x to the power of n

maka rumus integral aljabar diperoleh:

integral a times x to the power of n equals fraction numerator a over denominator left parenthesis n plus 1 right parenthesis end fraction x to the power of left parenthesis n plus 1 right parenthesis end exponent plus C

dengan syarat n not equal to 1.

Dengan memperhatikan rumus integral di atas, dapat ditentukan hasll dari integral fraction numerator left parenthesis x plus 1 right parenthesis left parenthesis x minus 3 right parenthesis over denominator square root of x end fraction d x yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 3 close parentheses over denominator square root of x end fraction d x end cell equals cell integral fraction numerator open parentheses x squared minus 2 x minus 3 close parentheses over denominator square root of x end fraction d x end cell row blank equals cell integral fraction numerator open parentheses x squared minus 2 x minus 3 close parentheses over denominator x to the power of 1 half end exponent end fraction d x end cell row blank equals cell integral open parentheses x to the power of 2 minus 1 half end exponent minus 2 x to the power of 1 minus 1 half end exponent minus 3 x to the power of 0 minus 1 half end exponent close parentheses d x end cell row blank equals cell integral open parentheses x to the power of 3 over 2 end exponent minus 2 x to the power of 1 half end exponent minus 3 x to the power of negative 1 half end exponent close parentheses d x end cell row blank equals cell 2 over 5 x to the power of 5 over 2 end exponent minus 4 over 3 x to the power of 3 over 2 end exponent minus 6 x to the power of 1 half end exponent plus C end cell row blank equals cell 2 over 5 x squared square root of x minus 4 over 3 x square root of x minus 6 square root of x plus C end cell end table

dimana C adalah konstanta. 

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 01 Agustus 2021

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