Question

Find:\[ \int \frac{dx}{(x + 2)(x^2 + 1)} \]

Answer 1

We use partial fraction decomposition:

\[ \frac{1}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1} \]

Multiply both sides by \( (x + 2)(x^2 + 1) \):

\[ 1 = A(x^2 + 1) + (Bx + C)(x + 2) \]

Expand both sides:

\[ 1 = A(x^2 + 1) + Bx(x + 2) + C(x + 2) \] \[ = A x^2 + A + Bx^2 + 2Bx + Cx + 2C \] \[ = (A + B)x^2 + (2B + C)x + (A + 2C) \]

Now, equate coefficients:

• \( A + B = 0 \)
• \( 2B + C = 0 \)
• \( A + 2C = 1 \)

Solving:

From \( A + B = 0 \Rightarrow B = -A \)

Substitute in \( 2B + C = 0 \Rightarrow 2(-A) + C = 0 \Rightarrow C = 2A \)

Substitute into third equation:

\[ A + 2(2A) = 1 \Rightarrow A + 4A = 1 \Rightarrow 5A = 1 \Rightarrow A = \frac{1}{5} \] \[ B = -\frac{1}{5},\quad C = \frac{2}{5} \]

Now rewrite the integral:

\[ \int \left( \frac{1}{5(x + 2)} + \frac{-\frac{1}{5}x + \frac{2}{5}}{x^2 + 1} \right) dx \] \[ = \frac{1}{5} \int \frac{1}{x + 2} dx - \frac{1}{5} \int \frac{x}{x^2 + 1} dx + \frac{2}{5} \int \frac{1}{x^2 + 1} dx \]

Final Integration:

• \( \int \frac{1}{x + 2} dx = \ln|x + 2| \)
• \( \int \frac{x}{x^2 + 1} dx = \frac{1}{2} \ln(x^2 + 1) \)
• \( \int \frac{1}{x^2 + 1} dx = \tan^{-1} x \)

Final Answer:

\[ \boxed{ \frac{1}{5} \ln|x + 2| - \frac{1}{10} \ln(x^2 + 1) + \frac{2}{5} \tan^{-1} x + C } \]