integral of 1/(2-x^2) — step-by-step solution

1) Let's integrate:

\int(\frac{1}{2 - x^{2}})dx

Let's factor in

ℝ

2 - x^{2}

Factors identified:

- (x + \sqrt{2}) \cdot (x - \sqrt{2})

2) You can decompose into partial fractions:

\frac{1}{- (x + \sqrt{2}) \cdot (x - \sqrt{2})} = \frac{A}{x + \sqrt{2}} + \frac{B}{x - \sqrt{2}}

For the simple factor:

x + \sqrt{2}

and its root:

- \sqrt{2}

, the unknown A can be easily computed by multiplying by the denominator:

2 - x^{2}

and evaluating at the root because the other terms cancel out:

3) Let's solve in

ℝ

- (- \sqrt{2} - \sqrt{2}) \cdot A = 1

- -2 \cdot \sqrt{2} \cdot A = 1

2 \cdot \sqrt{2} \cdot A = 1

A = \frac{1}{2 \cdot \sqrt{2}}

The conventional preference is to have numerical square roots in the numerator rather than in the denominator:

A = \frac{\sqrt{2}}{2 \cdot 2}

Solution obtained:

A = \frac{\sqrt{2}}{4}

For the simple factor:

x - \sqrt{2}

and its root:

\sqrt{2}

, the unknown B can be easily computed by multiplying by the denominator:

2 - x^{2}

and evaluating at the root because the other terms cancel out:

4) Let's solve in

ℝ

- (\sqrt{2} + \sqrt{2}) \cdot B = 1

- 2 \cdot \sqrt{2} \cdot B = 1

B = \frac{1}{- 2 \cdot \sqrt{2}}

B = - \frac{1}{2 \cdot \sqrt{2}}

The conventional preference is to have numerical square roots in the numerator rather than in the denominator:

B = - \frac{\sqrt{2}}{2 \cdot 2}

Solution obtained:

B = - \frac{\sqrt{2}}{4}

5) Finally, substitute the unknowns back into the initial partial fraction decomposition:

\frac{\frac{\sqrt{2}}{4}}{x + \sqrt{2}} - \frac{\frac{\sqrt{2}}{4}}{x - \sqrt{2}}

\frac{\sqrt{2}}{4} \cdot \frac{1}{x + \sqrt{2}} - \frac{\sqrt{2}}{4} \cdot \frac{1}{x - \sqrt{2}}

ℹ

\frac{\sqrt{2}}{4} \cdot \frac{1}{x + \sqrt{2}}

\frac{\sqrt{2}}{4 \cdot (x + \sqrt{2})}

‖

- \frac{\sqrt{2}}{4} \cdot \frac{1}{x - \sqrt{2}}

- \frac{\sqrt{2}}{4 \cdot (x - \sqrt{2})}

\frac{\sqrt{2}}{4 \cdot (x + \sqrt{2})} - \frac{\sqrt{2}}{4 \cdot (x - \sqrt{2})}

6) Let's integrate:

\int(\frac{\sqrt{2}}{4 \cdot (x + \sqrt{2})} - \frac{\sqrt{2}}{4 \cdot (x - \sqrt{2})})dx

\int(\frac{\sqrt{2}}{4 \cdot (x + \sqrt{2})})dx - \int(\frac{\sqrt{2}}{4 \cdot (x - \sqrt{2})})dx

Let

u = 4 x + 4 \cdot \sqrt{2}

. Then

du = 4 dx

, so

dx = \frac{du}{4}

\int(\frac{\sqrt{2}}{4 \cdot (x + \sqrt{2})})dx = \frac{\sqrt{2}}{4} \cdot \int(\frac{1}{u})du

We use

\int(\frac{1}{u})du = \ln (abs (u))

. The absolute value covers both signs of the denominator, on any interval where

4 x + 4 \cdot \sqrt{2}!= 0

Substitute

u

4 x + 4 \cdot \sqrt{2}

back into the antiderivative

Let

u = 4 x - 4 \cdot \sqrt{2}