Teka teki mpls penghalus kucing
Answer (3)
∫ ( x − 1 ) ( x 2 + 9 ) 10 d x = ( ∗ ) ( x − 1 ) ( x 2 + 9 ) 10 = x − 1 A + x 2 + 9 B x + C = ( x − 1 ) ( x 2 + 9 ) A ( x 2 + 9 ) + ( B x + C ) ( x − 1 ) = ( x − 1 ) ( x 2 + 9 ) A x 2 + 9 A + B x 2 − B x + C x − C = ( x − 1 ) ( x 2 + 9 ) ( A + B ) x 2 + ( − B + C ) x + ( 9 A − C ) ⇕ 10 = ( A + B ) x 2 + ( − B + C ) x + ( 9 A − C ) ⇕ A + B = 0 an d − B + C = 0 an d 9 A − C = 10 A = − B an d C = B → 9 ( − B ) − B = 10 → − 10 B = 10 → B = − 1 A = − ( − 1 ) = 1 an d C = − 1
( ∗ ) = ∫ ( x − 1 1 + x 2 + 9 − x − 1 ) d x = ∫ ( x − 1 1 − x 2 + 9 x + 1 ) d x = ∫ x − 1 1 d x − ∫ x 2 + 9 x + 1 d x = ∫ x − 1 1 − ∫ x 2 + 9 x d x − ∫ x 2 + 9 1 d x = ( ∗ ∗ ) #1 ∫ x − 1 1 d x ⇒ x − 1 = t d x = d t ⇒ ∫ t 1 d t = l n t + C 1 = l n ( x − 1 ) + C 1
#2 ∫ x 2 + 9 x d x ⇒ x 2 + 9 = u 2 x d x = d u x d x = 2 1 d u ⇒ ∫ ( 2 1 ⋅ u 1 ) d u = 2 1 ∫ u 1 d u = 2 1 l n ( u ) + C 2 = 2 1 l n ( x 2 + 9 ) + C 2
#3 ∫ x 2 + 9 1 d x = ∫ x 2 + 3 2 1 d x = 3 1 t a n − 1 ( 3 x ) + C 3 t h ere f ore : #1 ; #2 ; #3 ⇒ ( ∗ ∗ ) = l n ( x − 1 ) + C 1 − 2 1 l n ( x 2 + 9 ) + C 2 − 3 1 t a n − 1 ( 3 x ) + C 3
= l n ( x − 1 ) − 2 1 l n ( x 2 + 9 ) − 3 1 t a n − 1 ( 3 x ) + C
To evaluate ∫ ( x − 1 ) ( x 2 + 9 ) 10 d x , we decompose the integrand into partial fractions, from which we obtain terms to integrate separately. The resulting integrals give us the final expression ln ∣ x − 1∣ − 2 1 ln ∣ x 2 + 9∣ − 3 1 tan − 1 ( 3 x ) + C . This approach utilizes basic integration techniques and the properties of logarithms and arctangents.
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Jawaban:"Sisir"Alasannya: Sisir digunakan untuk menghaluskan bulu kucing.
