Question

∫ de la -1 la 3 1\x+2 dx

Answer 1

Răspuns:

I=∫

0

1

x

2

−16

1

dx=?

\begin{gathered}\it \dfrac{1}{x^2-16} = \dfrac{1}{x^2-4^2} = \dfrac{1}{(x-4)(x+4)} =\dfrac{A}{x-4} + \dfrac{B}{x+4} = \\\;\\ \\\;\\ = \dfrac{Ax+4A+Bx-4B}{(x-4)(x+4)} = \dfrac{(A+B)x +4(A-B)}{(x-4)(x+4)} \Longrightarrow \end{gathered}

x

2

−16

1

=

x

2

−4

2

1

=

(x−4)(x+4)

1

=

x−4

A

+

x+4

B

=

=

(x−4)(x+4)

Ax+4A+Bx−4B

=

(x−4)(x+4)

(A+B)x+4(A−B)

⟹

\begin{gathered}\it \Longrightarrow \begin{cases}\it A+B=0 \Rightarrow A= -B\ \ \ (*) \\\;\\ \it 4(A-B) =1 \Rightarrow A-B = \dfrac{1}{4} \stackrel{(*)}{\Longrightarrow} -B-B=\dfrac{1}{4}\Rightarrow B = -\dfrac{1}{8}\end{cases} \\\;\\ \\\;\\ \Longrightarrow A = \dfrac{1}{8}\end{gathered}

⟹

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

A+B=0⇒A=−B (∗)

4(A−B)=1⇒A−B=

4

1

⟹

(∗)

−B−B=

4

1

⇒B=−

8

1

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